
Class 

Book ^V 

7 

GopyiightN? 



COPYRIGHT DEPOSIT; 



MATHEMATICAL TEXTS 

Edited by Percey F. Smith, Ph.D. 

Professor of Mathematics in the Sheffield Scientific School 
of Yale University 



Elements of the Differential and Integral Calculus 
By W. A. Granville, Ph.D. 

Elements of Analytic Geometry 

By P. F. Smith and A. S. Gale, Ph.D. 

Introduction to Analytic Geometry 

By P. F. Smith and A. S. Gale, Ph.D. 

Advanced Algebra 

By H. E. Hawkes, Ph.D. 

Text-Book on the Strength of Materials 

By S. E. Slocum, Ph.D., and E. L. Hancock, M.Sc. 

Problems in the Strength of Materials 

By William Kent Shepard, Ph.D. 

Plane Trigonometry and Four-Place Tables of Logarithms 
By W. A. Granville, Ph.D. 

Plane and Spherical Trigonometry and Four-Place 
Tables of Logarithms 

By W. A. Granville, Ph.D. 

Four-Place Tables of Logarithms 

By W. A. Granville, Ph.D. 

Theoretical Mechanics 

By P. F. Smith and W. R. Longley, Ph.D. 



THEORETICAL MECHANICS 



BY 



PERCEY F. SMITH, Ph.D. 

PEOFESSOR OF MATHEMATICS IN THE SHEFFIELD SCIENTIFIC SCHOOL 
YALE UNIVERSITY 



AND 



WILLIAM RAYMOND LONGLEY, Ph.D. 

ASSISTANT PROFESSOR OF MATHEMATICS IN THE SHEFFIELD SCIENTIFIC SCHOOL 
YALE UNIVERSITY 



GINN AND COMPANY 

BOSTON • NEW YORK • CHICAGO • LONDON 



o 



> , 






Entered at Stationers' Hall 



Copyright, 1910, by 
Percey F. Smith and William Raymond Longley 

ALL RIGHTS RESERVED 

910.3 



GINN AND COMPANY • PRO- 
PRIETORS • BOSTON • U.S.A. 



©GI.A259978 



PREFACE 

The study of Mechanics as presented in this volume is founded 
upon a course in mathematics extending through the Calculus. 
It is assumed, moreover, that the student has already become 
familiar with the fundamental ideas of force, energy, and work 
through such preliminary courses as are included in textbooks 
on General Physics. In short, this volume presents the subject 
of Mechanics in that relation to other mathematical subjects which 
has become established in the curricula of the technical schools 
of this country. It should be emphasized, however, that the 
volume includes, for purposes of review, a discussion of the 
fundamental notions and many simple exercises involving these 
notions. 

Attention may be called to the arrangement in the text. This 
arrangement is founded upon experience in teaching the subject 
for many years in the Sheffield Scientific School of Yale Univer- 
sity. In 1903 Professor E. R. Hedrick prepared a mimeographed 
text which followed the conventional arrangement of treating 
statics first. This text was used for one year. It then developed 
that an obvious disadvantage existed in not taking up directly 
upon the conclusion of the study of the Integral Calculus the 
calculation of the integrals of Mechanics involving centers of 
gravity and moments of inertia. The point was that this formal 
integration out of the way, the continuous study of Mechanics 
proper need not afterwards be interrupted. Acting upon this 
conviction, the present text was prepared essentially as here 
published in 1907, and has since that time been used in mimeo- 
graphed form. The general plan of the arrangement is that 
a single problem may at any one time be under discussion. Thus, 
when the question of energy of rotation is solved, the appearance 
of the moment of inertia integral presents no complication. This 
has been disposed of already. Similarly, the equations of motion 
presenting themselves as solutions of the force equations have 



vi THEORETICAL MECHANICS 

been previously discussed. Another feature is the departure 
from convention by arranging types of motion under the corre- 
sponding fields of force. In this way it is made clear that the 
emphasis is to be laid upon the force and velocity of projection. 

In the case of a book which, like the present volume, has been 
long in the making, it is difficult to record definite acknowledg- 
ments of aid and indebtedness. There ar^e included in the text 
many problems suggested by past and present members of the 
mathematical department of the Sheffield Scientific School. 
Further, the text has been the subject of discussion at frequent 
departmental conferences, and for all suggestions received on 
these occasions the authors gratefully here record their thanks. 
The diagrams were skillfully prepared by Mr. S. J. Berard of 
the department of mechanical engineering. 

New Haven, Connecticut 



CONTENTS 



CHAPTER I. MOMENTS OF MASS AND INERTIA 

SECTION PAGE 

1. Center of gravity 1 

2. Moment of area 1 

3. Symmetry 3 

4. Theorem on the center of gravity 5 

5. Moment of mass (solids of revolution) 7 

6. Moment of mass (particular solids) 9 

7. Moment of mass. Any solid 12 

8. Principle of combination 14 

9. Center of gravity of an arc 17 

10. Theorems of Pappus . .18 

11. Moment of inertia. Plane areas 21 

12. Theorems on moments of inertia 22 

13. Further theorems 24 

14. Polar moment of inertia 24 

15. Flat thin plates or laminae . . .25 

16. Solids of revolution 26 

17. Moments of inertia of solids in general 28 

18. Parallel axes 30 

19. Relation between moment of inertia of a beam and polar moment 

of a right section 32 

20. Combined solids and areas 33 

21. Routh's rules 34 

22. System of material particles 36 

23. Ellipse of inertia 37 

CHAPTER II. KINEMATICS OF A POINT. RECTILINEAR MOTION 

24. Motion on a straight line 42 

25. Velocity 43 

26. Acceleration . • 45 

27. Distance-time diagram. Discussion 46 

CHAPTER III. KINEMATICS OF A POINT. CURVILINEAR MOTION 

28. Position in a plane or in space. Vectors 55 

29. Addition of vectors 56 

30. Subtraction of vectors 57 

vii 



viii THEORETICAL MECHANICS 

SECTION PAGE 

31. Multiplication of a vector by a scalar 57 

32. Resolution of plane vectors .58 

33. Vectors in space . . . 60 

34. Displacement in a plane. Path .64 

35. Velocity in the plane. Velocity curve 65 

36. Acceleration in a plane 69 

37. Motion in space \ .... 70 

38. Discussion of any motion 70 

39. Motion in a prescribed path 76 

40. Tangential and normal accelerations . . . . . .80 

41. Equations in polar coordinates 83 

42. Rotation 87 

CHAPTER IV. KINETICS OF A MATERIAL PARTICLE 

43. Momentum ............ 90 

44. Force 90 

45. Units of force 92 

46. Rectilinear motion . . .93 

47. Resultant force in rectilinear motion . 95 

48. Curvilinear motion. Axioms on force action. Concurrent forces . 100 

49. Curvilinear motion 101 

50. Intrinsic force equations . 105 

51. Polar equations 105 

CHAPTER V. WORK, ENERGY, IMPULSE 

52. Work . . . . . .109 

53. Kinetic energy . . 113 

54. Constrained motion. Dynamic pressure 116 

55. Units of work and energy. Power 122 

56. Impulse . 125 

57. Impact 127 

58. Force-moments in a plane 130 

59. Moment of momentum 132 

60. Angular momentum . 133 

61. Fundamental equations . 134 

62. Formulas in dynamics of a particle . 135 

CHAPTER VI. MOTION OF A PARTICLE IN A CONSTANT FIELD ■ 

63. Field of force 137 

64. Rectilinear motion under a constant force 137 

65. Curvilinear free motion 142 

66. Constrained motion 148 

67. Inclined plane . 149 

68. Motion on a smooth circle. Simple pendulum 154 



CONTENTS ix 

SECTION PAGE 

69. Motion on a smooth cycloid . . . . , „ . . 158 

70. Seconds pendulum 160 

CHAPTER VII. CENTRAL FORCES 

71. Central field of force 163 

72. Areal velocity 164 

73. Law of areas for central forces 165 

74. Converse of the theorem of areas 166 

75. The energy equation .167 

76. Circular' orbits 168 

77. Differential equation of the orbit 170 

78. Determination of the orbit when the law of force is known . . 173 

79. Position in the orbit 176 

80. Complete solution of a problem in central motion . . . .177 

81. Planetary motion. The law of gravitation 182 

CHAPTER VIII. HARMONIC FIELD 

82. Harmonic central field 186 

83. Energy equation 188 

84. Simple harmonic motion 188 

85. Composition of simple harmonic motions in a given field . . . 193 

86. Composition with different periods. Forced vibrations . . . 196 

87. General harmonic field 198 

CHAPTER IX. MOTION IN A RESISTING MEDIUM 

88. Law of resistance 200 

89. Constant field. Resistance proportional to the square of the velocity 200 

90. Damped harmonic motion. Resistance varying as velocity . . 202 

CHAPTER X. POTENTIAL AND POTENTIAL ENERGY 

91. Potential 205 

92. Conservative field . . 205 

93. Potential energy. Conservation of energy 206 

94. Equipotential lines and lines of force 207 

95. Non-conservative forces. Friction 213 

96. Newtonian potential .213 



CHAPTER XL SYSTEM OF MATERIAL PARTICLES 



97. System in a plane 

98. System in space 

99. Moment equation for a system of particles 

100. Work and energy of the system . 

101. Rigid system of particles . 



217 
218 
220 
221 
221 



THEOEETICAL MECHANICS 



CHAPTER XII. DYNAMICS OF A RIGID BODY 

SECTION PAGE 

102. Rigid body 226 

103. Translation 226 

101. Rotation 227 

105. Uniplanar motion ■ . . . . 227 

106. Centrodes . . . . . . > 229 

107. Screw motion 229 

108. Force equations. Work and energy . . . . . . . 231 

109. Kinetic energy 232 

110. Moment equation in rotation 234 

111. Comparison of formulas in translation and rotation .... 236 

112. Fundamental equations for uniplanar motion 236 

CHAPTER XIII. EQUILIBRIUM OF COPLANAR FORCES 

113. Equilibrium of forces 245 

114. Analytic conditions for equilibrium of coplanar forces . . . 245 

115. General method of solving problems in equilibrium .... 249 

116. Friction . . . .254 

117. Equilibrium of flexible cords 258 

118. The common catenary . 260 

119. Load distributed uniformly along the horizontal .... 261 

120. Stability 262 

CHAPTER XIV. COLLECTION OF FORMULAS 

Formulas from algebra 267 

Formulas from geometry • . . . 268 

Formulas from trigonometry . . 268 

Formulas from analytic geometry 269 

Formulas from calculus . . 271 

Differential equations 272 

Formulas for differentiation . . . 274 

Table of integrals 276 

Index . . . . .287 



THEORETICAL MECHANICS 



CHAPTER I 

MOMENTS OF MASS AND INERTIA 

1. Center of gravity. It is shown in a subsequent chapter 
(Art. 108) that the influence of the weight of a solid in all 
questions iri mechanics is precisely that of a force equal to the 
weight applied at a point called the center of gravity* of the solid. 
It is assumed that the student is familiar with simple facts con- 
cerning the center of gravity. For example, the center of gravity 
of a straight line (or thin straight rod) is its middle point. 
Again, the center of gravity of a triangle is the point of inter- 
section of the medians. 

A 
This statement may be proved as follows. Divide the 
triangle into thin strips by lines parallel to one side. Draw 
the median AD. The center of gravity of each strip lies on 
AD. Hence the center of gravity of the triangle lies on AD. 
Similarly, the center of gravity lies on the median BE. This 
establishes the statement. 

The formulas for the center of gravity in- _g 
troduced in the following sections involve mag- 
nitudes called the moments of area or moments of mass. The 
student is asked to accept these formulas as definitions. Later, 
in discussing weight the formulas appear as giving the center of 
gravity. 

2. Moment of area. Consider an element of any plane area 

y 

A A = AzAy, 

at the point (x, y). Then the products 









\ft 






/ 


t' \ 


/ y 




1 / 









X VAX 

y\ 



xAA, yAA 



x 

are called the moments of AA with respect to the axes OY and 
OX, respectively. 

* Called also center of mass. 
1 



THEOEETICAL MECHANICS 




This definition is extended to any finite area 
A in the usual way by summation and taking 
limits. Hence if M x and M y denote the mo- 
ments of area for the area A with respect to the 
X axes OX and OY, respectively, then 

f limit 

Mx = J $ ydA = j* §V di)C dy = j Ax = o, 2} 2) ybacbv 

m I A 2> = O, 

k ' C limit 

Jf y = f \xdA = \ (xdxdy - Ax = O, ^^ajAajAy 

. Ay = O, 

The Center of Gravity of any given area A is the point (#, y) 
given by the quotients 



(H) 



x=M- 
area 



area 



y = 



area 



area 



In these formulas a; and ?/ are the coordinates of any point within 
the area. 

The common denominator (the area of the given figure) must 
be found, if not otherwise known, by integration; that is, 

Area =11 dxdy. 

In working out examples using (II) calculate the moments 

j I xdA and / / yd A, first, and then divide by the area itself. 

Dimensions. Whenever it is desirable to express numerically 
the magnitude of a physical quantity, we do so by choosing a 
unit of that quantity. It is convenient, when possible, to choose 
the units of different kinds of quantities so that some of them 
depend upon others. The units which are chosen arbitrarily are 
called fundamental. The derived units are those which are so 
defined as to depend upon the fundamental units. In mechanics 
it is customary to choose as fundamental the units of length, mass, 
and time, and all other units are made to depend upon these. 
For example, if the unit of length is the foot, the unit of area 
is defined as the area of a square whose sides are one foot in 
length. The relation between the derived unit of area and the 



MOMENTS OF MASS AND INERTIA 



fundamental unit of length is then expressed by the dimensional 
equation, Area = i engt h2. 

Similarly, the dimensional relation between the derived unit of 
volume and the fundamental unit of length is 

Volume = length 3 . 

The dimensional equation is a concise way of expressing the 
relation between the units of different quantities, and is not to 
be interpreted as an ordinary algebraic equation. 

Moment of area has been defined as the product of area by 
distance, and hence the unit of moment of area is of the third 
degree in the unit of length. 

Moment of area = area x length = length 3 . 
The fact that every term of an equation involving physical 
quantities must be of the same degree in the fundamental units 
furnishes a useful check in the problems of mechanics. For 
example, in (II) x is of the first degree in the unit of length, and 
hence the second member of the first equation must also be of the 
first degree. This is easily verified, since the dimensional relation 



gives 



M y 'length* 



area length 



length. 



3. Symmetry. The center of gravity will lie upon any axis of 
symmetry which the figure may possess. For ex- 
ample, if OY is such an axis, we may divide the 
figure into the equal elements AxAy and sum up, 
taking two symmetrical pairs at a time. Then the 
sum of the moments with respect to OY for two 
such pairs, that is, x 1 AxAi/ + x 2 AxAi/, will vanish, 
since x x — — x 2 . Hence the mo- 
ment with respect to OY, that is, 

My =11 xdA, also vanishes, and x — 0. 

Illustrative Example. Find the center of gravity 
X of the area bounded by y 1 = 2 px and x = h. 
Solution. Evidently y — 0. 

Calculate the moment of area with respect to OY 
This is, by (I), 

My = \ \ xdxdy = \x\ dydx — 2 V2p \x*dx = %V2p h 2 ' 





4 THEOEETICAL MECHANICS 

Next find the area. This is 

r n +s/2 PX 



dydx = 

h. by (ii). 



PROBLEMS 

Note. If the equation of the curve is given in polar coordinates (p, 0), place 
in (I) and (II) dA = pdpdd, x = p cos 0, y =p sin 6. 

~ 1. Find the center of gravity of the triangle bounded 

J^ by the lines y = mx, y=0,x = a. A ^-2 n j_»« 
— 'pcos'Sy^T ^ ns - x ~ 3 a ' V - ~Y - 

y/^ \psinO 2. Find the center of gravity of the triangle bounded 

/^W L__ b y the lines y = mx, y = -mx,y = b. Ans. x = 0, y= f b, 

Ji. 

3. Find the center of gravity (1) of a quarter of a 

circle in the first quadrant ; (2) of one sixth of a circle, supposing the z-axis to be 
an axis of symmetry. ^ (1) - = - = 4a . (2) - = 2« - Q 

4. Find the center of gravity of a quadrant of the ellipse — _i_ ^_ = i, 

a' 2 b 2 

Ans. x = ^,y=**. 

5. Find the center of gravity for the area bounded by y 2 = 4 ax, y — 0, x — b. 

Ans. x = f&, y = iVab. 

6. Find the center of gravity of the area bounded by y 2 = 4 ax, x = 0, y = b. 

3 h2 _ q 

Ans. x— ,y = -b- 

40a 4 

7. Find the center of gravity of the area bounded by the semicubical parabola 
ay 2 = x^ndx=a. ^ - =M< 

8. Find the center of gravity of the area bounded by y = asm- and the 
cc-axis between x = and x = air. 



a 



Ans. x=l air, 



9. Find the center of gravity of the area bounded by the hyperbola xy = e 2 , 

x = a, x = 6, and y = 0. Ang - _ b-a ? - _ c 2 (ft - a) _ 

log 5 — log a 2 a& (log 6 — log a) 

10. Find the center of gravity of the area bounded by the parabola y 2 = 4 ax 

and the straight line y = mx. _ q a _ o /> 

^4ws. x=— — , ?/ = 

5 m 2 m 

11. Find the center of gravity of the area included by the curves y 2 = ax 
and x z = by. A - q L| _ 21 

12. Find the center of gravity of the area bounded by the cardioid 

p = a (1 + cos 6) . ^4ws. 5 = | a. 






MOMENTS OF MASS AND INERTIA 5 

13. Find the center of gravity of the area included by a loop of the curve 

P = a cos 2 0. _ 128 a V2 
Ans. x = 

105 TV 

14. Find the center of gravity of the area included by a loop of the curve 

p:=acos3 0. _ 81aV3 

15. The lengths of the parallel sides of a trapezium are a and b. Show that 
the center of gravity of the area divides the line joining the middle points of the 
parallel sides in the ratio (a + 2 6)/(2 a + 6). 

16. If the sides of a triangle be 3, 4, and 5 feet, find the distance of the center 
of gravity from each side. Ans. f, 1, f foot. 

17. Find the center of gravity of the area bounded by the cissoid 

y 2 (2 a - x) = cc 3 
and its asymptote x = 2a. Ans. x = f a. 

18. Find the center of gravity of the area bounded by the witch 

x 2 y = 4 a 2 (2 a — y) 
and the axis of X. Ans. y — \a. 

19. Find the center of gravity of the area bounded by the parabolas y' 2 = ax 
and y 2 = 2 ax — x 2 , which is above the axis of X 

Ans. x=a l5ir ~ U ;y 



15 7T _ 40 3 7T - 8 

20. Find the distance from the center of the circle of the center of gravity of 

the area of a circular sector of angle 2 d. . „ sin 8 

° Ans. |r . 

9 

21. Find the distance from the center of the circle of the center of gravity of 
the area of a circular segment, the chord subtending an angle 2 6. 

Ans. ? rsin ' e 



3 6 — sin0cos<? 



4. Theorem on the center of gravity. The center of gravity 
of an area is & fixed point relative to that area. That is, the posi- 
tion of the center of gravity does not de- 
pend upon the axes of coordinates, but 
upon the area itself only. The proof of 
this familiar truth is as follows. 

Let L be any line, and assume its equa- 
tion in the normal form (55 (e), Chap. XIV) 

x cos w -f- y sin &> — p = 0. 

Consider the element of area AA= AxAy at (#, #), and let the 
distance from L to (#, y) equal r. Then the product rAxAy is 




THEOEETICAL MECHANICS 



called the moment of AA with respect to the line L. Extending 
to a finite area as before, the double integral 



(1) 



M L = C CrdA 



is called the moment * of area with respect to L. 

This integral may be expressed in terms of the moments M z 
and M y with respect to OX and OF as follows. By Analytic 
Geometry,f or formula 56, Chapter XIV, 

r = x cos co + y sin co — p. 
.-. I lrdA= I I (#cos co + y sin co — p)dA 

= cos col I xdA + sin col lydA—pi f dA 

= cos co My + sin coM x — pA. 
Using formulas (II), putting A = Area of the figure, then 



My = Ax, M x = Ay, 



If 



dA—A. Hence 



M L = (x cos co + y sin co — p)A = rA, 

if r = distance from L to the center of gravity (x, y). 
Hence this 

Theorem. The moment of area of a plane figure with respect 
to any line equals the product of the area and the distance from that 
line to the center of gravity. Hence the moment of area with respect 

to any line through the center of gravity 

r is zero. 

Now suppose we have worked out 
the coordinates of the center of gravity 
C for a plane figure with respect to a 
given set of axes OX and OY. Let 
0' X, 0' Y' be any other set of axes. 

Let the new coordinates of any point in the area be (V, y f }. 

Also let the new coordinates of (7 be {xJ , y f ). Then, by Art. 2, 

* Also called the first moment, because of the appearance of the first power of the 
distance r in the integral. 

t Smith and Gale, Elements of Analytic Geometry (Ginn and Company), p. 106. 
Future references are to this volume. 




MOMENTS OF MASS AND INERTIA 



formulas (II), the coordinates of the center of gravity found by 
using the new axes are 

C Cx'dA C Cy'dA 

area area 

By the theorem just given we have, however, 

I I x'dA = moment of area with respect to OY' = Ax', 



fj 



y'dA = moment of area with respect to OX' = Ay' . 

f r r . , . r r . • i 



(V, y') ; that is, the same cen- 



C Cx'dA C Cy'dA 

Hence I J-d. , «z_*z 

1 A A 

ter of gravity is found by using the new axes. This investiga- 
tion, therefore, verifies a well-known property of the center of 
gravity, namely, that it is a fixed point relative to the area. 

SOLIDS OF REVOLUTION 

5. Moment of mass. The volume of a thin flat plate or lamina 
equals the product of its surface by the thickness. If of uniform 
density, its mass is the product of the volume and the density. 
For the present, the density will be assumed constant and will be 
denoted by r. The lamina being thin, its center of gravity is 
sensibly the same point as the center of gravity of its surface or 
area. The moment of mass of a lamina with respect to a plane 
parallel to its surface equals the product of its mass and the dis- 
tance from the plane to its surface. The plane being parallel to 
the surface of the lamina, every point of the lamina is at the same 

2 Z 




(a) (b) (c) 

distance from the plane. Passing now to a homogeneous (of 
uniform density = t) solid of revolution, we may slice up such a 
solid by a series of equidistant parallel planes perpendicular to 
the axis of revolution (fig. a). Assume OX as this axis, and Ax 



8 THEORETICAL MECHANICS 

as the common thickness of the slices. Now consider each slice 
" trimmed up " into a circular lamina, one face of the slice remain- 
ing unchanged, so that the solid is now replaced by a new solid 
obtained by revolution of the set of rectangles in fig. b. The 
mass Am of any one of the circular laminae is (fig. c) 

Am = tAv = t • 7ry ? Ax, 

for y( = z) is the radius of the base and Ax the thickness. Since 
the lamina is parallel to YZ, its moment of mass with respect to 
YZ is xAm or t • iry 2 Ax times x. The total moment of mass of 
all the circular laminw may then be represented by *2xAm or also 
^Tiry^xAx. The moment of mass of the solid itself is then 
defined as the limiting value of this sum when Ax approaches zero. 
Using for this the symbol M ye , we have 

(III) M ys = (ocdm = tit \xy 2 doci = ^ _ q VrTrai^Aa? j. 

The method explained here of slicing the solid of revolution 
into circular laminae is very important and should be mastered by 
the student. 

The center of gravity of a solid of revolution whose axis is 
along OX is defined as the point (#, y, z), where 



-*r j xdm m I xy 2 dx 



(IV) ^=^£-=^ - = — ^ »-£ = 0, 5 = 0. 

v y mass mass mass 

It is clear that y = z = 0, since the centers of gravity of all the 
laminae are on the axis of revolution, and hence the center of 
gravity of the solid is on the axis of revolution. 

In the calculation of x, we need to find two integrals, 

Myg = tit I xy 2 dx and Mass = / dm = tit I * 

in which y is to be found in terms of x from the equation of the 
generating curve. 

Dimensions. The quantity moment of mass has been defined as 
the product of mass by distance. Hence in terms of the funda- 
mental units of mass and of length the dimensional relation is 

Moment of mass = mass x length. 



MOMENTS OF MASS AND INERTIA 9 

PROBLEMS 

HOMOGENEOUS SOLIDS OF REVOLUTION 

1. Find the center of gravity of the cone formed by revolving the line hy = ax 
around the re-axis between x = and x = h. Ans. x = f h. 

2. Find the center of gravity of a hemisphere. 

Ans. Distance from base = f radius. 

3. Find the center of gravity of the paraboloid of revolution formed by revolv- 
ing about the sc-axis the parabola y 2 = 4 ax from x = to x = b. 

Ans. x = f b. 

4. The area bounded by the lines y = 0, x = a and the curve y 2 = 4 ax is 
revolved about the ?/-axis. Find the center of gravity of the solid formed. 

Ans. y = 4 a. 
x 2 y 2 

5. The area of the ellipse — ^ + ^ = 1, in the first quadrant, is revolved about 

the x-axis. Find the center of gravity of the solid formed. Ans. x : = f a. 

x 2 v 2 

6. The area bounded by the lines y = 0, x = 2 a and the hyperbola — , — ~ = 1 

is revolved about the x-axis. Find the center of gravity of the solid formed. 

7. The area bounded by the lines x = 0, x = a, y = 0, and the hyperbola 

x 2 y 2 

— = — ^hs + 1 = is revolved about the x-axis. Find the center of gravity of the solid 
a 2 b 2 

formed. 

8. The area bounded by the lines y = 0, x — — , and the curve y = sin x is re- 
volved about the x-axis. Find the center of gravity of the solid formed. 

9. The area bounded by the lines x = 0, x = a, y = 0, and the curve y = e x is 
revolved about the z-axis.' Find the center of gravity of the solid formed. 

10. Find the center of gravity of the solid generated by a semiparabola 
bounded by the latus rectum, revolving round the latus rectum. 

Ans. Distance from focus = -g- 5 ^ °f latus rectum. 

PARTICULAR SOLIDS 

6. Moment of mass. Certain solids may be divided by a series 
of parallel planes into laminae whose sur- 
faces depend in a simple manner only upon 
their distances from a parallel fixed plane. 
Taking this plane as YZ and considering a 
lamina at the distance x, then if A is its 
surface, by hypothesis, A=f(x), — a known 
function. Hence Aw = rf(x) Ax (if the 
thickness of the lamina is Ax~). The moment of mass of the 
solid with respect to YZ will then be defined as equal to 




10 THEORETICAL MECHANICS 

(1) M yz = fxdm = rfx f(x)dx(= l ™\ 2ra/(3>AA 

The distance x of the center of gravity from the PZ-plane is of 
course equal to /» 

M r l x f(x)dx 

(2) x=-^- = ^l ___, 

V J Mass /' , ' 



cfo 



since 



Mass equals / dm = r f f(x)dx. 



In (2), the uniform density r cancels out. The function 
f(x), it is to be remembered, is the area of a cross section parallel 
to TZ at the distance x. 

Illustrative Example. Find the center of gravity of any cone, pyramid, 
or cylinder of uniform density. 

Solution. The definition of a cone or pyramid 

must be clearly understood. This is the following. 

Given any plane area B and a point V without it. 

Draw the line VP through Fand any point P on the 

boundary of the area P. Now let the point P move 

around the boundary of P, carrying in its motion the 

line VP. The surface thus generated by the line VP, 

called a generator, and the area B bounds a solid. 

If B is bounded by straight lines, the solid is a 

pyramid, otherwise a cone. The area B is called the 

base and V the vertex. 

The following theorem is now assumed for any cone or pyramid. Take a section 

A parallel to the base B. Then the areas of A and B are in the same ratio as the 

squares of their distances from the vertex V. 

To apply formula (2) , let the area B lie in the FZ-plane. Let the section A be 
at the distance x from the base. Draw the line VH perpendicular to the base P, 
and let VH = h = altitude. Then 

distance of the area A from vertex = h — x, 
distance of the area B from vertex = h. 

.-. by the theorem, - = ( h ~ x ) 2 ri = -(/i- x) 2 . 
B h 2 h 2 ^ J 

Hence in (2), f( x )=-(h- x) 2 . 
h 2 

.-. M yz = r( — 7 x(h- x) 2 dx = -L rBh 2 , 

h- 




M=t( - y (h - x) 2 dx = i rBh. 
Hence x = \h. 




MOMENTS OF MASS AND INEBTIA 11 

Now it is clear that the centers of gravity of all sections of the cone or pyramid 
which are parallel to the base B will lie on a line joining Fto the center of gravity 
of the base. This line is called the axis. Hence the 

Theorem. The center of gravity of any homogeneous cone or pyramid is the 
point on the axis which is one fourth of the distance from the 
base to the vertex. 

A cylinder is the solid obtained thus. Let a generating 
line AA' move always parallel to itself, while the point A 
follows a plane curve inclosing an area B. The solid 
bounded by this surface, by the area B, and by the section B' 
parallel to B, is called a cylinder. The line joining the cen- 
ters of gravity of B and B' is called the axis. This line is parallel to the generator 
A A'. Clearly, the center of gravity of the cylinder is the middle point of the 
axis. 

PROBLEMS 

1. Find the center of gravity of a frustum of a pyramid with a square base. 

2. Find the center of gravity of an elliptic cone. The equation of an elliptic 

v 2 z 2 
cone is *— A — = x 2 . Take the plane x = 1 for the base of the cone, 
a? b 2 

3. Find the center of gravity of the solid bounded by the elliptic paraboloid 

— J r ^- = z, and the plane = 1. 
a 2 b 2 

4.* Find the center of gravity of a right conoid with circular base, the radius of 
the base being r and altitude a. 

5. A rectangle moves from a fixed point, one side varying as the distance from 
this point, and the other as the square of this, distance. Find the center of gravity 
of the solid generated while the rectangle moves a distance of 2 feet. 

x 2 v 2 

6. On the double ordinates of the ellipse — -f *— =1, isosceles triangles of verti- 

a 2 b 2 

cal angle 90° are described in planes perpendicular to that of the ellipse. Find the 
center of gravity of the solid generated by supposing such a variable triangle moving 
from one extremity to the other of the major axis of the ellipse. 

7. Given a right circular cylinder of altitude a and radius of base r. Through a 
diameter of the upper base pass two planes, which touch the lower base on opposite 
sides. Find the center of gravity of the solid included between the planes. 

8. Two cylinders of equal altitude a have a circle of radius r for their common 
upper base. Their lower bases are tangent to each other. Find the center of 
gravity of the solid common to the two cylinders. 

9. An anchor ring is cut in two equal parts by a plane through its center, which 
passes through its axis. Find the center of gravity of one half. 

*For the volumes of the solids of examples 4-8, see Granville, Differential and 
Integral Calculus (Ginn and Company), p. 422. Future references are to this volume. 



12 



THEOEETICAL MECHANICS 



7. Moment of mass. Any solid. Consider any solid and an 
interior point (#, y, z). The density of this solid may be variable. 
In this case, we assume the density t at any interior point (x, y, z) 

to be some function of the co- 
ordinates, say 

(1) density at (x, y, z) 

= r(x,y,z). 

Taking an element of vol- 
ume 

Av=AxAy As, 

we have as the element of 
mass at (x, y, 3), 

Am = t (#, y, z) Av. 

The moment of mass for this element with respect to the coordi- 
nate planes we define thus : 

with respect to YZ=x- Am, 
" " " ZX=yAm, 

" " " XY= z- Am. 

The moments of mass of the solid with respect to the coordinate 
planes are derived from these by summation and passing to the 
limit as Ax, Ay, and Az approach zero. That is, we define for 
any solid, 

(V) Mzy = \ \ \xdm, M zx = \ \ Kydm, M xy = \ \ \zdm, 




the limits being so chosen that the entire solid is included, 
mulas (V) are included in the single formula 



For- 



M-- 



SIS' 



rdm, 



where r is the distance from one of the coordinate planes to any 
interior point of the solid. In these formulas x, y, and z are the 
coordinates of any point within the solid. The center of gravity 
of the solid is then the point whose coordinates aj, y, z are given 

by 



(VI) 



- M v 



oc 



mass 



M zx 
mass 



z = 



M xv 
mass 



MOMENTS OF MASS AND INERTIA 13 

In formulas (V) and (VI) we set 

dm = r (#, y, z) dxdydz, mass = I j j dm. 

In determining the center of gravity of a solid, four integrals, 
namely, the moments with respect to the three coordinate planes 
and the mass, must be calculated. 

Homogeneous solids. In this case the density t is constant. 
For such solids, a theorem corresponding to that of Art. 3 holds, 
namely, 

The center of mass of a homogeneous solid lies in any plane of 
symmetry of the solid. The proof is left to the reader. To derive 
formula (III) (Art. 5) from (V), proceed thus. We have 



M yz — T / I j xdxdydz — T I \ I I dydi 
Bat If dydz \= area of cross section in the plane x = constant, 



xdx. 

x= constant 



x= constant 



and hence equals Try 2 under the conditions of Art. 5. 

. *. My Z — T I 7ry 2 xdx, 
which is (III). 

Theorem on the Center of Mass. Results analogous to 
those of Art. 4 are readily derived for solids ; namely, 

The moment of mass of a solid with respect to any plane equals 
the product of the mass by the distance from the plane to the center 
of gravity. The center of gravity is a fixed point relative to the solid. 
This proof is left to the reader. 

PROBLEMS 

1. Find the center of gravity of the first octant of the homogeneous ellipsoid 

x2 , v 2 z 2 i 

~ + t-s + -£= 1 ' Ans. x= f a, y = %b, z = f c. 



2. Find the center of gravity of the homogeneous solid bounded by the surface 
z* = xy, and the planes x = a, y = b,z = 0. An§ - = f ^ - = f ^ - = ^ ^ 

3. Find the center of gravity of the paraboloid of revolution formed by revolv- 
ing about the z-axis the parabola y' 2 = 4 ax from x = to x = b, supposing the den- 
sity to vary as x 2 . Ans. x=%b. 



14 THEORETICAL MECHANICS 

4. Find the center of gravity of a hemisphere whose density varies as x 2 , as- 
suming the base in the TZ-plane and the origin at the center of the base. 

Ans. x = I a. 

5. Find the center of gravity of the cone formed by revolving the line hy = ax 
around the x-axis between x = and x = h, assuming the density varies as x n . 

Ans. x^l+JLh. 

n + 4 

6. Find the center of gravity of the homogeneous solid bounded by the surfaces 
x 2 + y 2 = 4 2, a 2 + y 2 = 3 x and z = 0. 

7. The axes'of two cylinders each of radius a intersect perpendicularly. Find 
the center of gravity of the solid included by the two cylinders and a plane through 
their axes. Ans. f a from the plane. 

8. A thin plate whose density varies as (h 2 — x 2 )~* is bounded by the lines 
y = ax, y = 0, and x = a. Find its center of gravity. Ans. x= \ irh ; y = \ irah. 

9. Find the center of gravity of the first quadrant of a circular plate whose 
density varies as xy. Ans. x = y = ^a. 

10. Find the center of gravity of a circular sector (angle = 2 6, radius = a) if 

the density varies as the distance from the center. . - _ Set sin 6 

x _ __ . __ 

11. Find the center of gravity of a circular sector in which the density varies 

as the nth power of the distance from the center. 

n+2 ac 
Ans. — —ft • t, where a is the radius of the circle, I the length of the arc, and 

c the length of the chord of the sector. 

12. Find the center of gravity of a circle in which the density at any point 
varies as the ?zth power of the distance from a given point on the circumference. 

Ans. It is on the diameter passing through the given point at a distance from 

2 (n + 2) 
this point equal to — , 4 • a, a being the radius. 

13. Find the center of gravity of a quadrant of an ellipse in which the density 
at any point varies as the distance of the point from the major axis. 

Ans. x =-a, y = — - b. 
8 '* 16 

8. Principle of combination. Since the moment of area or of 
mass is a definite integral, if an area or solid 
is divided into two parts, the moment of the 
s^ whole equals the sum of the moments of 
4A the separate parts. Thus consider the ac- 
~j 2 J companying figure, in which (x v y-^ is the 
y^ center of gravity of the area m A v and (x 2 , ?/ 2 ) 

— i the center of gravity of the area A 2 . Tak- 
ing moments with respect to 
OX: total moment = Ay = A 1 y 1 -\- A 2 y 2 ; 
OY: total moment = Ax= A x x x 4- A 2 x 2 . 



MOMENTS OF MASS AND INERTIA 15 

Hence the center of gravity of the combined areas is 

^ il) X ~ A 1 + A 2 ' y ~ A 1 + A 2 ■ 

These formulas agree with those for the point of division in 

A 

formula 50, Chapter XIV, if X =~- Hence this 

A i 

Theorem. The center of gravity of a plane figure composed of 
two parts divides the line joining the centers of gravity of the parts 
in the inverse ratio of the areas of the parts. 

A similar theorem holds for solids. 

The discussion holds for an area (or solid) resulting when a 
portion of the area (or solid) is removed, if its area or mass be 
taken negatively. The proof, which is left to the reader, comes 
from (VII) by transposition. In working problems under this 
head, the line joining the centers of the parts may conveniently 
be taken for one axis of coordinates. 

Illustrative Example. To find the center of gravity of the remainder of a 
circle of radius 2 r after a circle of radius r has been removed as indicated in the 
figure. 

Solution. Let c be the center of gravity sought, 
and denote the area of the large circle by A 2 and 
that of the small circle by A\. Then we have 

A 1 = -irr^ 
A-2 = 4 7rr 2 . 

Substituting in (VII), 

-_ — irr 2 ■ r + 4 irr 2 • _ r 

Evidently y is zero by symmetry. Hence the center of gravity c lies on the £-axis 
at a distance of ^ r to the left of the origin. Also c divides the line Cic 2 in the ratio 
> A 2 




= -4. 
A x 



PROBLEMS 



1. A rod of uniform thickness is made up of equal lengths of three substances, 
the densities of which taken in order are in the proportion of 1, 2, and 3; find the 
position of the center of mass of the rod. 

A?is. At T ?g of the whole length from the end of the densest part. 

2. If five ninths be cut away from a triangle by a line parallel to the base, 
show that the center of gravity of the remaining area divides the median in the 
ratio 4:5. 



16 



THEORETICAL MECHANICS 



3. One corner of a square plate of side a is cut off by a line joining the middle 
points of two adjacent sides. Find the center of gravity of the remainder. 

Ans. from the center. 

21 

4. An equilateral triangle is formed on one side of a square. Find the center 



of gravity of the whole area. 



Ans. 



3a 



from base of triangle. 



8 + 2V3 

5. One corner of a square of side 2 a is cut off by a line drawn from a corner 
to the middle point of an opposite side. The opposite corner is also cut off by 
removing a circle of radius p having its center at the corner. Find the center of 
gravity of the remainder. 

6. Find the centers of gravity of the shaded portions of the following figures. 

€11 



^&p 




7. A cylinder is 12 in. long, and for 8 in. of its length has a diameter of 4 in.; 
for the remaining 4 in. it has a diameter of 3 in. Find the center of gravity. 

Ans. 5|f in. from thick end. 



MOMENTS OF MASS AND INERTIA 17 

8. A cone having the same base and vertex is cut from the paraboloid of revo- 
lution whose generating curve is y' z = 4 ax between x = and x = b. Find the 
center of gravity of the remaining solid. . - _ b 

2 

9. From a sphere of radius B is removed a sphere of radius r, the distance 
"between their centers being c. Find the center of gravity of the remainder. 

Ans. It is on the line joining their centers and at a distance — — — from the 

7?3 _ r 3 

center. 

10. Find the center of gravity of a cubical box without a lid, the inside edge 
being 20 in. and the thickness of the wood 1 in. 

11. Find the center of gravity of the remainder of an equilateral triangle from 
which has been cut an isosceles right triangle with hypotenuse coincident with a 
side of the original triangle. 

12. A right circular cone whose base is of radius r is divided into two equal 
parts by a plane through the axis. Prove that the distance of the center of gravity 

v 

of either half from the axis is - • 

7T 

13. Find the center of gravity of half of a regular hexagon. 

14. From a hemisphere is cut a cone having the same base and altitude. Find 
the center of gravity of the remainder. Ans. Distance from base = \ altitude. 

15. From a right circular cylinder is cut a cone having the same base and 
altitude. Find the center of gravity of the remainder. 

Ans. Distance from base = f altitude. 

16. From a right circular cone of altitude a is cut a similar cone of altitude b, 
the bases of the two cones being in the same plane. Find the center of gravity of 
the remainder. ^ Bist ^ ce from base = 1 a? - b\ 

4 a s — b 3 

9. Center of gravity of an arc. The center of gravity for any 
plane curve is given by formulas analogous to (II), Art. 2, ob- 
tained by replacing the element of area or mass by the element of 
arc of the curve, that is, for a plane curve, by 66, Chapter XIV, 



1 + 



dx 



[■ + £§)'] 



dy. 



(1) A = [(<&)» + 0*y)»]' 
The formulas are y 

(VIII) S = J^! vJ VdS 

arc ' J arc 

in which ds is found by (1). In these x and y 

are the coordinates of any point on the curve. 

Formulas (VIII) are used to find the center of gravity of uni- 
form thin wires. If a is the area of the cross section, and As the 



y 



18 



THEOEETICAL MECHANICS 



length of a piece whose projections on OX and OY are Ax and 
Ay, respectively, then for the mass of this piece we write 

Am = a As times the density (= t) ; 
(1) or Am = t • crAs. 

For the moments of mass with respect to 
OX and OY of this piece, we have the 
— y products 

yAm and xAm, 

respectively. Thus we obtain for a plane-curve wire as in 
formulas (II), 




(2) 



mass 






I rcryds I i 
fas J, 



since the constant, o-, divides out. If the wire is uniform, t is 
also constant, divides out, and we have (VIII). 



Illustrative Example. Find the center of gravity of a quadrant of the 

n 



2 2 2 

hypocycloid x 3 + y 3 = a 3 . 



Solution. Consider the part of the curve in the first 
quadrant. 
Then 



j" «b = ( x Vl + (|) 2 ft = aijj xkx = ? * 

[my 



J^ = j*o^U) 2 + 1 ^ = f" 2 




f ds = ( -\ll + ( -£ ) die = - a. Hence, applying (VIII), x = y = | a 



dx 



10. Theorems of Pappus. Consider any area in the JTZ-plane. 
The distance of the center of gravity from the x-'dxis is given 
by the formula (II), 



MOMENTS OF MASS AND INEBTIA 



19 



where A denotes the area. Let the area be revolved about the 
#-axis. The volume generated is given by the definite integral 

(2) V: 



= rr I y 2 dx. 



Integrat- 



V 



Consider the numerator in (1). 
ing with respect to y, 

(3) / ydydx = - Jy 2 dx = - ^ 

comparing with (2). 

Substituting in the second member of (1), we get 




y 



i y a o - y 



Now 2 Try = length of path described by the center of gravity. 
Hence the 

First Theorem. If any plane area be revolved about an ex- 
terior axis in its plane, the length of the path described by its center 
of gravity is equal to the volume generated,- divided by the area 
revolved. 

This theorem has two uses: (1) if the area and its center of 
gravity are known, we may find the vol- 
ume of the solid of revolution; (2) if the 
area and volume are known, we may find 
the center of gravity. For example, to 
find the distance of the center of gravity 




of a semicircle from the center, we have 



2rry = 



volume of sphere i7r« 3 



_ ±a 



semicircle 



2 , whence y = ~~. 



Next consider any curve in the XY- 

plane. The distance of the center of 

gravity from the a>axis is given by the 

formula (VIII), /• 

/ yds 
(5) y = d , 

s 

where s denotes the length of the curve. Let the curve be re- 
volved about the a>axis. The surface generated is (68, Chap. XIV) 

S =2tt f yds. 




20 THEORETICAL MECHANICS 

But the length of the path described by the center of arc is 
(multiplying both members of (5) by 2 7r) 



2tt yds „ 

V = 2Z _ _. 



s S 

Hence the 

Second Theorem. If any plane curve be revolved about an ex- 
terior axis in its plane, the length of the path described by its center 
of gravity is equal to the surface of the solid generated, divided by the 
length of the arc revolved. 

This theorem has two uses: (1) if 

mthe length of the arc and its center are 
known, we may find the surface of the 
solid of revolution; (2) if the length of 
the arc and the surface of the solid are 
known, we may find the center of gravity of the arc. 

For example, to find the distance of the center of gravity of a 
semicircle from the center, we have 

o - surface of sphere 4 ira 2, -, — 2 

Airy — — f- = , whence y = —a. 

semicircumference ira it 



PROBLEMS 

1. Find the center of gravity of an arc of the circle p = a between — 6 and + 0, 
and from this derive the results for quadrantal and semicircular arcs. 

Ans. x = a sm d - For quadrantal arc = -, x~ 



4 7 wr 

For semicircular arc = —, x = 

2 7T 

2. Find the center of gravity of a thin straight wire of length a whose density 

varies as the nth power of the distance from one end. A - n + 1 

r Ans. x = a. 

n+2 

3. Find the center of gravity of the perimeter of the cardioid p = a (1 + cos 6). 

Ans. x = f a, y = 0. 

V I 

4. Find the center of gravity of the cycloid x = a arc vers — (2ay — y 2 )^ 

fj^Y 9/ \ ft 

between two successive cusps. Hint. — = — y Ans. x = air, y = — . 

dy V2ay-y* 3 

5. Find by the theorem of Pappus the center of gravity of one fourth of a circle 

in the first quadrant. A o -_-_4a 

ji.ns, x — y — — — • 

3 IT 



MOMENTS OF MASS AND INERTIA 



21 



6. Find by the theorems of Pappus the volume and surface of the torus gener- 
ated by revolving the circle (x — b) 2 + y 2 = a 2 (b>a) about the ?/-axis. 

Ans. V=2w 2 a 2 b, S = 4w 2 ab. 



7. The ellipse 



ifi 



1 is revolved about the line x = 2 a. Find by the 



theorem of Pappus the volume generated. 



Ans. 4 ir 2 a 2 b. 



8. An equilateral triangle revolves around its base, whose length is a. Find 
(1) the area of the surface and (2) the volume of the solid generated. 

Ans. (1) 7ra2V3; (2) ^. 
4 

9. A square of side a is revolved around an axis in its plane, the perpendicular 
distance of which from the center is c. Find (1) the area of the surface and (2) the 
volume of the solid generated. 

10. A rectangle is revolved around an axis, which lies in its plane and is per- 
pendicular to a diagonal at its extremity. Find the area of the surface and the 
volume of the solid generated. 



11. Moment of inertia. Plane areas, 
area 

AA = Ax Ay, 



Consider an element of 
Y 




at the point (x, y). The products, 

x*AA, y 2 AA, 

are called the moments of inertia or second 
moments of A A with respect to the axes OY and OX respec- 
tively. The definition is extended to a finite area by summation 
and passing to the limit. Using I x , I y for the moments of inertia 
with respect to OX and Z", respectively, then 



(IX) 



y 2 dA = J J y 2 dxdy = Aoc = O j>]y 2 AxAy 



Ay 
limit 



I y = f I x 2 dA = \ \x 2 dxdy = Age - O ^?ac 2 AxAy 



In these formulas x and y are the coordinates of any point within 
the area. Formulas (IX) are embraced in the single formula 



(i) 



■-// 



rHA, 



where r is the distance from the axis in question to any point 
within the area. This integral is called also the second moment 
of area, from the second power of the distance r. 



22 



THEOEETICAL MECHANICS 



Since each element y 2 AxAy or ofiAxAy is essentially posi- 
tive, the moment of inertia * is never zero, but a positive 
number. 

Its dimensions are area times square of a 
length, and hence it is of the fourth degree 
in the fundamental unit of length. 

Illustrative Example. Find I y for the portion 

of y 2 = 2 px cut off by x = h. 

Solution. We have, by (IX), 

W2px 
W2px 

i — 3 
Since A = ±V2ph 2 , we get for I y the expression 



Y 

i 


ff 


sT 


*>< 


AXy 


If 


4 






JL 




X 



Iy = j (tfdxdy = f T j*_ 2p \ly]x 2 dx = 2 V2^ Cx^dx = $V2ph\ 



SA 



h*. 



Note. 
in (IX) 



PROBLEMS 

If the equation of the curve is given in polar coordinates (p, 0), write 

d A = pdpdd, x = p cos 0, y = p sin 0. 

1. Eind /for a rectangle of sides 2 a and 2 6: (1) with respect to an axis through 
the center of gravity parallel to the side 2a; (2) with respect to the side 2 a. 

Ans. (1) — ; (2) -AW: 

2. Eind 7 for a circle with respect to a diameter. Ans. £ Aa 2 . 

3. Find I for an ellipse : (1) with respect to its major axis ; (2) with respect 
to its minor axis. Ans. (1) \ Ab 2 ; (2) \ Aa 2 . 

4. Find I for a right triangle with respect to one side. 

5. Find / for a square with respect to a diagonal. Ans. -^ Aa 2 . 

6. Find /for an equilateral triangle with respect to a median. 

7. Find I x for the cardioid p = a(\ + cos 0). 



8. Find I x and I y for one loop of the curve p = a cos 2 0. 

9. Find I y for the lemniscate p 2 = a 2 cos 2 0. ylras. 7 y 



48 



(3 ir + 8) a 3 



12. Theorems on moments of inertia. The moment of inertia 
of the element of area AA = Ax Ay with re- y 
spect to any line or axis L equals 

r*AA, 

where r is the distance from the line L to 
the point (x, y). The moment of inertia of 
a finite area with respect to L is then 




* It appears later that moment of inertia determines the kinetic energy of revolution. 




MOMENTS OF MASS AND INEBTIA 23 

(X) II = J jjr 2 dA = tj §r 2 dxdy, 

in which r is the perpendicular distance from the line L to any 
point (x, y) within the area. 

Let us apply (X) to the case of an axis parallel to OX, whose 
equation is y = a. Then r — y — a, and hence 

II = ff(y - afdA =ff(f -2ay + <fi)dA 
= C Cy 2 dA -2a f CydA + a 2 f CdA. 

(1) .-. J z =£-2«Jf,+AA(by(rc) 

Art. 11, and (II) Art. 2). 

This formula expresses the moment of in- 
ertia I L in terms of the moment of inertia 
with respect to any parallel axis OX, the 
moment of area with respect to the latter, 
and the area itself. 

But suppose the center of gravity lies on OX. Then y = 
and also M x = 0. Hence 

(XI) I L = I x +a*A. 

An axis passing through the center of gravity is called a grav- 
ity axis. 

This establishes the important 

Theorem. The moment of inertia of a plane area with respect 
to any axis equals the moment of inertia with respect to the parallel 
gravity axis, increased by the product of the area by the square of the 
distance between the axes. 

This statement shows that the moment of inertia with respect 
to a gravity axis is less than the moment of inertia for any parallel 
axis. 

Radius of gyration. The quotient of the moment of inertia by 
the area is the square of a length called the radius of gyration. 
Thus, if r L denote this, 

in which r L is the radius of gyration with respect to the axis L. 



24 THEORETICAL MECHANICS 

PROBLEMS 

1. Find the radius of gyration in the problems on page 22. 

2. Find / and r for a circle with respect to a tangent. Ans. I = f Aa 2 , r 2 =| a 2 . 

3. Find /and r for an ellipse with respect to a tangent (1) at the end of the 
major axis ; (2) at the end of the minor axis. Ans. (1) / = f Aa 2 ; (2) / = f Ab 2 . 

4. Find /for a right triangle with respect to a line through one vertex parallel 
to the opposite side. 

5. Find /for a square with respect to a line through one vertex parallel to the 
diagonal joining the other two vertices. 

6. Find / for an equilateral triangle with respect to a line through one vertex 
parallel to a median. 

13. Further theorems. In the preceding section, the axis L 
was drawn in the plane of the given area. It is necessary, however, 

to consider moments of inertia with re- 
spect to axes without, but parallel to this 
plane. Let L be such an axis in the fig- 
ure. Then if r is the perpendicular dis- 
tance from the axis L to any point (x, y') 
within the area, we define in a manner 
precisely analogous to the foregoing, 




: C Cr*dA= C Cr 2 dxdy. 



Now project the line L upon the plane of the area, and take this 
projection as the axis OX. Let the distance between L and 
OX equal a. Then evidently r 2 = a 2 + j/ 2 , and hence 

I L = C |V + a*)dA = f CfdA + a 2 C CdA. 

(XII) .-. I L = I x + a 2 A. 

The moment of inertia of an area with respect to an axis 
parallel to its plane equals the moment of inertia with respect to the 
projection of the given axis on its plane increased by the product of 
the area by the square of the distance from the axis to the plane. 

14. Polar moment of inertia. The moment of inertia of an 
area with respect to the origin is defined as equal to 

(XIII) I = § 1j(oc 2 + y 2 )dA=§ §oc 2 dA-]-1j §y 2 dA. 



MOMENTS OF MASS AND INERTIA 



25 



It will be observed that (x 2 + y 2 ) A A is the product of A A by 
the square of the distance from (#, y) to an axis through 0, per- 
pendicular to the plane of the area. y 

Such an axis is called a polar axis. 

Comparison with (IX), Art. 11, enables 
us to write (XIII) in the form 

(XIV) I Q =I X + I r 

Hence the 

Theorem. The moment of inertia of an area with respect to a 
polar axis (called the polar moment) equals the sum of the moments 
with respect to two mutually perpendicular axes 
drawn through its foot. 

If polar coordinates (p, 0) are used, the 
origin being the pole, I , the polar moment of 
inertia, is given directly by 





(XV) Jo = j* | p 2 • pdpdQ = j" 1 9'dpdQ. 



Moments of inertia of a circle. On account of important appli- 
cations in the next section, the moments of inertia of a circle are 
now worked out. 

Let a = radius. Then, by (XV), the 
polar moment of inertia with respect to 
an axis through the center is 



a) i«= 



Jo LJo 



dO 



p s dp 



ira* 



= —a' 




2 
where A = area of the circle. 

Also since I x = I y , by symmetry, we have, by (XIV), 

(2) 



I —~I ——a 2 



In words: the polar moment of inertia of a circle with respect to 
its center equals the product of one half the area and the square of 
the radius; with respect to any diameter — the product of one fourth 
the area and the square of the radius. 

15. Flat thin plates or laminae. Moments of inertia of 
laminae are obtained from the corresponding moments of inertia of 
their surfaces by replacing the area by the mass of the lamina. 



26 



THEORETICAL MECHANICS 



For example, the polar moment of inertia of a circular lamina with 
respect to its center equals the product of one half the mass by the 
square of its radius. 
D 





TJie moment of inertia of a circular lamina with respect to a 
diameter equals the product of one fourth its mass by the square of 
the radius. 

16. Solids of revolution. Moments of inertia of such solids 
are obtained by slicing and trimming the solid into circular laminae 

by a series of equidistant 
planes perpendicular to the 
axis of the surface, and con- 
sidering the limit of the 
sum of the moments of 
inertia of the laminae. If 
the axis of revolution be 
chosen as OX, the common 
thickness of the laminae as A x, and the density as t, the mass A m 
of any lamina is 

(1) A m = T7n/ 2 A x. 

Moment of inertia of a solid of revolution with respect to the axis 
of revolution. The moment of inertia of any one lamina with 
respect to the axis of revolution is the same 
as the polar moment of a circular lamina 
with respect to its center. By Art. 15, 
this moment is equal to 

(2) Am 
2 




^V = ^VA* by(l). 




= 2 

The moment of inertia of the solid is accordingly 

(XVI) I* = ±tjy*dm=$^dx, 

in which y is to be found in terms of x from the equation of the 
generating curve. 



MOMENTS OF MASS AND INERTIA 



27 



Illustrative Example. Find the moment of inertia with respect to the axis 
of revolution of a cone formed by revolving about the x-axis the line y = mx between 
x = and x = b. 

Solution. From (XVI), 

2 Jo 10 

Since the radius of the base a = mb and the volume =^^ — , we have 
I x = & Ma 2 . 3 



PROBLEMS 



1. Find /for a rectangle of sides 2 a and 2 6 with respect to a line perpendicular 
to the plane and passing through the center. , T— —( 2 4. ?2n 

o 

2. Find /for a right triangle with respect to a line perpendicular to its plane 
and passing through the vertex of the right angle. 

Ans. I Ac 2 , where c is the hypotenuse. 

3. Find / for the area of an ellipse with respect to an axis perpendicular to 
the area and passing through the center. Ans. 1= \ A{a 2 + b 2 ). 

4. Find /and r for a sphere with respect to a diameter. Ans. 1= § Ma 2 . 

5. Find I x and f x for an ellipsoid of revolution about the x-axis. 

Ans. I=%Mb 2 - 

6. Find / and f for a right cylinder with respect to its axis. 

Ans. 1= \ Ma 2 . 

7. Find I x for the solids of revolution about the x-axis whose generating 
curves are ( a ) y 2 = ±ax f rom x = to x=6 

(5) ?/ = sin a; " as = " x = tt\ 

(c) y = mx + 6 " x = " x = c 

(d) y = e* " x = " x = a. 

Moment of inertia of a solid of revolution with respect to an axis 
cutting the axis of revolution at right angles. We wish to find the 
moment of inertia with respect to OY. 
To do this, we must first find the moment 
of inertia of one lamina with respect 
to OY. Now OY is an axis parallel to 
the surface of the lamina. Let DD f be 
the projection of OY upon this surface. 
Then, by (XII), Art. 13, for the lamina 
we have 

(") -Ly (for one lamina) = -*-D + # Am. 

But I D is the moment of inertia of the lamina with respect to 
a diameter. Hence, by Art. 15, 

Am 9 




(4) 



Id = 



28 THEOEETICAL MECHANICS 

Substituting in (3) gives 

(5) ly (for one lamina) == , V ~T" x AW*. 

Summation and passing to the limit leads to the result 

(XVII) Iy (for the solid) = /7~ + oAj dm = T / (^ + aA TTy 2 dX, 

in which y must be expressed in terms of x from the equation of 
the generating curve. 

PROBLEMS 

1. Find the moment of inertia of a right cylinder of radius a and altitude h 
with respect to a diameter of the base. . J— — (3 a 2 4- 4 A 2 ^ 

2. Find the moment of inertia of a right circular cone of altitude h and radius 
of base a, with respect to an axis through its vertex and perpendicular to its geo- 
metrical axis. Ans. I = ^ M (4 h 2 + a 2 ). 

3. Find the moment of inertia of the cone of problem 2 with respect to a 
gravity axis perpendicular to its geometrical axis. Ans. I = -^ M (h 2 -f 4 a 2 ). 

4. Find I y for the solids of problem 7, p. 27. 

17. Moments of inertia of solids in general. Consider any- 
solid and an interior point (#, y, z). If the density at this point 
is t (x, y, z) (compare Art. 7), the element of mass is 

(1) Am = TAxAyAz. 

The moments of inertia of Am relative to the coordinate planes 
are defined as 

(2) I yz = x 2 Am, I zx — y 2 Am, I xy = z 2 Am. 

The square of the distance of (x, y, z) from the axis of x being 
y 2 -f- z 2 (with similar expressions for the other axes), the moments 
of inertia of Am with respect to the coordinate axes are 

(3) I x = (y 2 + z 2 )Am, I y = (x 2 + z 2 ) Am, I 2 = (x 2 + y 2 )Am. 

The moments of inertia for the entire solid may now be written 
down, namely, 



(XVIII) 



(XIX) 



=555 



oc 2 dm 



limit 



^Va? 2 Am 
Ay = 0^ 



Izx = 55 1" v 2dm ' ix * = 5 5 5 zdm ; 

I x = j* § §(y* + z*)dm, I y = j* j" j\z 2 +x*)dm, 
I* = j* j* fj(x 2 + y 2 )dm, 



MOMENTS OF MASS AND INERTIA 29 

where dm = t(x, y, z) dxdydz, and (x, y, z) is any point within 
the solid. 

Formulas (XVIII) and (XIX) are included in the formula 



Iff- 



r 2 dm, 



where r is the perpendicular distance from the axis or plane in 
question to any point within the solid. 

Dimensions. The moment of inertia of a solid has been denned 
as the product of mass by the square of the distance. Hence the 
derived unit of moment of inertia is expressed in terms of the 
fundamental units of mass and of distance by the dimensional 
equation 

Moment of inertia = mass x length 2 . 

By the radius of gyration of a solid with respect to any axis is 
understood a length r t whose square is the quotient of the moment 
of inertia with respect to the axis by the mass. Thus 

(3a) r 2 = -i-, etc. 

mass 

The relations 

\*J -*-x = -*-zx ~T~ J-xyi -*-y == -*-yz ~T~ -L x y> ^z == -*-yz ~T J-zxl 

obviously hold. In words, 

The moment of inertia of a solid with respect to any axis equals 
the sum of its moments relative to two mutually perpendicular planes 
passing through the axis. 

Homogeneous solids. For such solids the density r is every- 
where constant. Formulas (XIX) applied in this case to a homo- 
geneous solid of revolution about the #-axis work out as follows : 



dx. 

x— constant 



(5) I x=T fff(f + ^dxdydz = t f\_ff(f + *Yh'l* 

x — 

But / l(y 2 + z 2 )dydz calculated for any plane section x = 

constant is obviously the polar moment of a circle with respect to 
its center. Since the radius of this circle is y, then (Art. 14) 

(6) ffif + * 2 ) dydz = 3? . f = Qt 



•so 



THEORETICAL MECHANICS 



(7) 



- 1 *-/ 1 



dx. 

x — constant 



which is (XVI), Art. 16. 

Similarly for the same solid, 

(8) I y =r C C i\x* + z 2 )dxdydz 
= t I \ If dydz \x 2 dx + t I II z 2 dydz \ 

x= constant x = cons 

But / l (^/cte calculated for any plane section when x = con- 
stant, is the area of that section; that is, 

(9) jjdydP^wf. 

/» /* a; = constant 

Again, I I z 2 dydz for the section # = constant is the moment of 

inertia of that section with respect to its diameter in the plane 
XT, Hence (Art. 14), 

(10) jj z 2 dyd ^ = ^f. y 2 t 

Substituting in (8) gives 

(11) I y = r Cirx 2 y 2 dx + r f ^ <fe ; 
that is, (XVII). 



18. Parallel axes. If U is any plane, the moment of inertia of 
any solid with respect to U is defined as 

lE= SSJ r2dw ' 

where r is the perpendicular distance from 
the plane to any point (#, y, z~) within the 
solid. 
Parallel planes. Let U and E' be 
two parallel planes, r and r' the dis- 
tance from them to any interior point 
(#, y, z) of a solid. Then if a is the 
common distance apart of E and E\ 
we have 

r = r r + a, and hence 








MOMENTS OF MASS AND INERTIA 
I £ = f C f r 2 dm = f f C(r' + dfdm 
= C C Cr'Wm +2a C C Cr'dm + a 2 f f fdm. 
i f f fr' 2 dm = I E ', f f fdm = M, 



31 



r' 2 dm 

and III r'dm = Mr, where r is the distance of the center of 
gravity of the solid from E' , Hence 

(1) I E = I £ >+2 aMr + a 2 M. 

Suppose E' passes through the center of mass. Then f = 0, and 
we have the important result 

(2) I E = I E < + am. 

Any plane passing through the center of a mass is called a grav- 
ity plane. 

Theorem. The moment of inertia with respect to any plane is 
equal to the moment of inertia with respect to the parallel gravity 
plane, increased by the product of the entire mass and the square of 
the distance between the planes. 

Since in any set of parallel planes one and only one passes 
through the center of the mass, it follows at once from (2) that 
of all moments of inertia with respect to parallel planes that with 
respect to the gravity plane is the least. 

Parallel axes. Let L and L' be any two parallel lines. Let E ' ' 
be the plane passed through the two lines L and L r , and let E and 
E be planes through L and L\ respectively, 
perpendicular to E" '. Then, from (1), 

I E = I £ i + 2aMr+a 2 M. 

Adding I E " to both members, we have 

(3) I E + I E " = Igi + I E " + 2a Mr + a 2 M. 
But, by (4), Art. 17, 

I E + I E i = I L and I E i + I E " = II > 
Also if L' (and consequently E f } passes through the center of 
mass, we have r = 0, and (3) becomes 

(4) I L = I L > + a 2 M. 




32 



THEORETICAL MECHANICS 



Hence the theorem stated above holds when the word " plane " is 
replaced by "axis." 

PROBLEMS 

1. Derive formulas for the moments of inertia of plane arcs (or wires). 

Ans. I x = ( ry 2 ds; I y = I rx 2 ds ; ds = (dx 2 + dy 2 )*- 

2. Find I for a solid cylinder with respect to an element. Ans. f Ma 2 . 

3. Find / for a solid sphere with respect to a tangent line. Ans. \ Ma 2 . 

4. Find I for a solid ellipsoid of semiaxes a, b, c with respect to the axis a ; 
with respect to a tangent line at the extremity of the axis b. 

Ans. M*±*; M 6 -*±*. 
5 5 

5. Find /for a uniform wire in the form of an equilateral triangle of side «, 
(1) with respect to a line perpendicular to the plane of the triangle and equidistant 
from the vertices ; (2) with respect to a line through a vertex perpendicular to the 
plane. ^ (1) Jftf 

v 2 

6. Find /for a solid cylinder with respect to a line perpendicular to its axis and 
intersecting it at a distance c from the end, the altitude of the cylinder being h and 
the radius of the base c. Ans. \ Mc 2 + \ M(li 2 — Shc + Sc 2 ). 

7. Find /for a straight rod of length a with respect to an axis perpendicular 
to the rod and at a distance d from its middle point. . tit (—4- d 2 \ 

8. Find / for an arc of a circle whose radius is a and which subtends an angle 
2 a at the center, (1) with respect to an axis through its center perpendicular to its 
plane ; (2) with respect to an axis through its middle point perpendicular to its plane ; 
(3) with respect to the diameter which bisects the arc. 

Ans. 



(1) Ma 2 ; (2) 2 M (l - ^ V; (3) M (l - ^~-)f 



9. Find / for the arc of the cycloid x = a (0 — sin 6) , 
respect to the base. 



a (1 — cos0) with 



19. Relation between moment of inertia of a beam and polar 
moment of a right section. Consider any homogeneous straight 
beam (density = t) whose elements are par- 



r i 



fCa 



W 



allel to OZ. Then, by (XIX), 
(1) I 2 = T C C C(x 2 + «/ 2 ) dxdydA 



-Jiff 



(x 2 + ?/ 2 ) dxdy 



dz. 

z= constant 



But / I (x 2 + y 2 ) dxdy, worked out for any section z = constant, 



MOMENTS OF MASS AND INERTIA 33 

is the polar moment of that section for the axis OZ (Art. 14). 
Hence (1) becomes 

(2) I z (for the beam) = ^ (for a right section) X height of Cylinder ( = Jl) X T. 

Let r s and r be the radii of gyration of beam and right sec- 
tion, respectively. Then 







r s 2 = — 2—, r 2 
mass 


area 


or 


also 






I.- Mr*, I = 


:r*A. 






Substituting 


in (2) gives 








(3) 




il!fr, 2 = 


Ar ViT. 






But M = 


Ahr, and hence 








(4) 




r 5 = 


-- V 







Theorem. The radius of gyration of any homogeneous beam with 
respect to an axis parallel to its elements equals the radius of gyration 
of a right section with respect to the same axis. 

From (4) we may write 

I z = Mr s \ I»=Ar s \ 

and hence the change from I to I s is. accomplished by replacing 
the area by the mass of the cylinder. In this form the result is 
useful and gives this 

Rule. To find the moment of inertia of a straight beam or 
column with respect to an axis parallel to its elements (or edges), 
work out the corresponding polar moment for any right section 
and replace in this result the area by the mass of the beam or 
column. 

20. Combined solids and areas. Since the moment of inertia is 
a definite integral, it follows that if a solid or area is composed of 
two or more parts, the moment of inertia of the whole with respect 
to any plane or axis is equal to the sum of the moments of inertia 
of its parts with respect to that plane or axis. Also, if a portion 
be removed from a solid or area, the moment of inertia of the 
remainder equals the moment of inertia of the whole minus the 
moment of inertia of the part removed. 

As an example, consider the polar moment of inertia with 
respect to its center of the circular ring formed by removing from 



34 



THEORETICAL MECHANICS 



a circle c of radius R, a concentric circle c' of radius r. Denoting 
the area of c by A, and that of c' by A\ the polar moment of inertia 
of c by Tq, and that of c' by j^', we have 
T - AR2 T , _A'r 2 

Hence the polar moment of inertia of the re- 
maining ring is 

1= \ (AR 2 - J/r 2 ) = £(i2 4 - r 4 ) 

= '?L(R2 + r 2 ) (U 2 -r 2 ). 
The area of the ring J[ is 

Hence 




A = nR 2 - 77T 2 . 



I = 



R 2 +r 2 ). 



That is, the polar moment of inertia with respect to its center 
of a circular ring lying between two concentric circles of radii R 
and r is equal to one half the product of its area by the sum of 
the squares of the radii. 

By the principle of Art. 19, we may at once extend this result 
to apply to a hollow circular column of outer radius R and inner 
radius r. Denoting by I the moment of inertia of the column 
with respect to its axis, we have 

I=^(R? + r 2 ). 

Theorem. The moment of inertia of a homogeneous hollow cir- 
cular column with respect to its axis is equal to one half the product 
of its mass by the sum of the squares of the inner and outer radii. 

21. Routh's rules. The following moments of inertia occur 
frequently and should be committed to memory: 

The moment of inertia of 

(1) a rectangle whose sides are 2 a and 2 b with ' 
respect to an axis through its center in its plane 
perpendicular to the side 2 a 

with respect to an axis through its center per- 
pendicular to its plane 



= M~ 



= M 



+ b 2 



MOMENTS OF MASS AND INERTIA 35 

(2) an ellipse of semiaxes a and b with respect | M - b 2 
to the major axis (a) J ~~ 4 ' 

a 2 
with respect to the minor axis (&) = M —r ; 

(a circle is an ellipse with semiaxes each equal to a) 

(3) an ellipsoid of semiaxes a, b, c, with respect ' 
to the axis (a) 

(a sphere is an ellipsoid with a = b = e) 

(4) a parallelopiped whose edges are 2 a, 2 5, 2 e, ' 
with respect to an axis through its center perpen- 
dicular to the plane containing the sides b and e 

(5) a circular cone the radius of whose base is a 1 3 
with respect to its axis J 10 

As an aid to the memory, the first four rules may be combined 
into one known as Routh's rule: 

/Sum of squares of per- 
— M - v Vpendicular semiaxes 



= jf 5 i±^ 



O 

Ma 2 . 



Moment of inertia 
with respect to an 
axis of symmetry 



3, 4, or 5 



9 The denominator is to be 3, 4, or 5, according as the body is 
rectangular, elliptical,, or ellipsoidal. 

As an example of the application of Routh's rule, suppose it is 
required to find the moment of inertia of a circle of radius a with 
respect to a diameter. We notice that the perpendicular semi- 
axis in its plane is a and the semiaxis perpendicular to its plane 

a 2 
is zero. Hence the moment of inertia is M — . Again, let it be 

4 

required to find the moment of inertia with respect to a line 

through the center of the circle and perpendicular to its plane. 

The perpendicular semiaxes are each equal to a and the moment 

of inertia is 

M a2 + a2 = M^. 



PROBLEMS 

1. Find the moment of inertia of the hollow column of Art. 20 with respect to 
a line perpendicular to the XF-plane, (1) through the outer circumference ; 
(2) through the inner circumference. Ans< ^ M ^ R2 ^ . M ^ R2 + 3 ^ 

2 2i 



36 



THEORETICAL MECHANICS 



2. Find the moment of inertia of the circular ring, Art. 20, relative to OX. 

3. Find the moment of inertia of the ring with respect to the tangents to the 
circles c and c'. 

4. Find the moment of inertia of a circular area having a smaller circular area 
cut from it as in the figure, (1) with respect to a line through perpendicular to 
the plane of the circle ; (2) with respect to a diameter of the larger circle perpen- 
dicular to 00'; (3) with respect to a line through 0' perpendicular to the plane of 

the circle ; (4) with respect to the diameter 00'. 




Ans. (1) ifJfi? 2 ; (2) 



(3) ff JfR2; 



(4) 



5. A square is removed from a circle, the diag- 
onals of the square intersecting at the center of the 
circle. Find I with respect to (1) an axis passing 
through the center of the circle perpendicular to its 
plane ; (2) an axis perpendicular to the plane and 
passing through one corner of the square ; (3) a diam- 
eter which is also a diagonal of the square. 

6. Find the moment of inertia with respect to the gravity axis parallel to an 
edge of the beams whose cross sections are shown in the following figures. 

2a' 
2b 



Zl^ 



(1) 





2a 


V 


i 


2b 


' 





3a 





22. System of material particles. By a material particle, or 
simply particle, is meant a portion of matter of so small a volume 
that the volume is regarded as reduced to a point. In other 
words, it is a weighted point or point mass. The moment of mass 
of a particle of mass m at the point P with respect to any line or 
plane equals the product of m by the perpendicular distance to P 
from the line or plane. 



MOMENTS OF MASS AND INERTIA 37 

The center of mass of any system of particles of mass m 1 at 
Pi(x v y v 2^), m 2 at P 2 (x 2 , y 2 , g 2 ), etc., is defined by the equations 

- m,x, -f- m 9 x -f- ... *Lmx 



(i) 



m 1 + w 2 + ••• 2m 

m 1 + 7n 2 -f- ••• 2/m 

- _ yw 1 2 1 + w 2 g 2 + •" _ ^ W3 



m x + m 2 + ••• Sm 

Similarly, the moment of inertia of a particle of mass m at P 
with respect to any axis equals the product of m by the square of 
the perpendicular distance from P to the axis. 

Thus for a system of particles lying in one plane whose masses 
are m 1 at P^x v y x ), m 2 at P 2 (# 2 , y 2 ), etc., we have 

^r = ™>\V\ + W 2^2 2 + ' • • = 2?W?/ 2 , 

I y = m^j 2 + w* 2 £ 2 2 -+-••• = 2ra# 2 , 
J = J r + /„ = 2m(^ 2 + */ 2 ) = 2 V- 

PROBLEMS 

1. Three edges of a unit cubical frame without weight are taken as the coordi- 
nate axes, and particles are placed at all the corners except at the origin. Find / 
with respect to each face, edge, and vertex of the cube, (1) when the particles are 
of equal mass ; (2) when the masses vary as the squares of their distances from the 
origin. 

2. A straight rod of negligible mass and length a has five particles of equal mass 
situated on it at equal intervals of \a. Find /and r ' 2 , (1) with respect to one end; 
(2) with respect to the middle point ; (3) find I when the masses increase in arith- 
metical progression from the end. 

3. Given three particles of equal mass, situated at the vertices of an equilateral 
triangle. Find (1) /and r 2 with respect to one side ; (2) with respect to a line 
parallel to one side passing through the opposite vertex. 

4. A regular hexagon has particles at middle points of five of its sides. The 
masses of the particles taken in order are as 1, 2, 3, 4, 5. Find /and r 2 with 
respect to the unweighted side. ^ nSm j _ 20.25 a 2 ; r 2 = 1.35 a 2 . 

23. Ellipse of inertia. This section is concerned with the solu- 
tion of the problem, 

To determine the moment of inertia of an area with respect to any 
gravity axis. Let be the center of mass of a given area ; OX, 



38 THEOBETICAL MECHANICS 

OY any two mutually perpendicular axes through it, and I 
any other gravity axis making with OX the angle 6. Then, by 

Art. 11, (1), 



¥ 








*-w\ 


^1 




/\o } / 




.^ 




* 



(1) *-// 



r 2 <M, 



where r is the perpendicular distance 
from I to any interior point (x, y) of 
the area. The equation of I may be 
written 

(2) — x sin + y cos = 0, 
and hence (56, Chapter XIV) we have 

(3) r = — x sin + y cos 0, 

when (x, y) is the interior point in question. Substituting in (1), 

(4) J ? = / /(-:zsm0 + y cos dydA, or, 

J z = sin 2 (9 T f z 2 ^ - 2 sin cos C CxydA + cos 2 f fy 2 dA. 

The second integral in the right-hand member has not thus far 
been discussed. If we set this equal to P xy , we may write 

(5) J z = I, cos 2 - 2 P xy sin cos + i" y sin 2 0, 
where 

(XX) 2%= f CxydA, 

and is called the product of inertia with respect to the axes OX 
and OY. It is easy to see that IJ assumes a maximum and a 
minimum value as the axis I rotates about O. In fact, since 

(6) -5= -27 x cos0sin0-2P^(cos 2 0-sin 2 0)+27 y sin0cos0, 

setting the right-hand member equal to zero gives 

(7) (/,-/,) sin 20-2P X2/ cos2 = 0, 

from which tan 2 = j* v T * 

-*-y -*-x 

The values of determined by this equation will give axes ? t and 

l 2 for which I t is a maximum and a minimum respectively. More- 



MOMENTS OF MASS AND INERTIA 



39 



over, since these values of differ by — , l x and Z 2 are perpendic- 
ular. They are called the principal axes of inertia. 

Obviously, if P xy — 0, the roots of (7) are — 0, = — , and 

A 

hence OX and OY &ve already the principal axes. Let us now 
assume this to be the case. Then (5) becomes 

(8) Ii=I x cos 2 + J^sin 2 ^. 

Introducing the radii of gyration by setting 

±i = Ar { , l x = ^lr x , ./y = .Ar^ , 

then (8) becomes 

(XXI) rf = r x 2 cos 2 6 + r,/ sin 2 6- 

This equation gives the radius of gyration with respect to any 
axis in terms of the principal radii of gyration, r x and r y . For 
convenience we now write 

(9) 



1 _1 

a ' b 

Thus (XXI) becomes 



(10) r t 



2_ 



cos 2 fl sin 2 6 



5 2 



Y 


— Sp^ 


*C l 


' r 


jf% 




\^ 


x J 


) x 


^ 







Let us now draw the ellipse, 

(ii) i 



t + tL 



a" b^ 

If (/>, 0) are the polar coordinates of the point P where the axis 
I cuts this ellipse, then in (11) 

x = p cos 0, y = p sin 0, and we get 
( 12) l = £^ + filler also 



1 _ cos 2 fl sin 2 
~f?~ a* b 2 ' 

Comparison with (10) gives the result 



(13) 



r 2 = 



The ellipse (11) is called the ellipse of inertia, and the result 
indicated by (13) may be stated thus: 



40 



THEORETICAL MECHANICS 



If the ellipse of inertia is drawn for any plane area, the radius 
of gyration for any gravity axis equals the reciprocal of the radius 
vector of the point in which the axis intersects the ellipse. 

The principal axes of inertia are those for which the product 
of inertia is zero ; that is, 



=11 xydxdy 



0. 




It is easy to see that P xy = if either OX or OYis an axis of 
symmetry. For example, if OX is such an axis, then in the sum 
of the products 

xyAxAy 

the terms will occur in pairs with the same 
x and with y\ differing only in sign. The 
terms in each such pair will cancel, and 
hence the limit of the sum is also zero. This 
consideration gives the result : 

Any axis of symmetry is necessarily a 
principal axis. 

The process, then, of determining the moment of inertia with 
respect to any gravity axis is the following : 

(1) If the figure has no axis of symmetry, choose any pair of 
rectangular axes, calculate I x and I y by (IX), and P xy by (XX). 
Then use equation (5), or solve equation (7) for 6 and determine 
the principal axes and the principal radii of gyration. Choose 
these axes for the new axes of coordinates and draw the ellipse of 
inertia (11). Then apply the theorem just stated to find r l or 
use (XXI). 

(2) If the figure has an axis of symmetry, choose this for 
OX or OY, calculate r x and r y , and draw 
the ellipse of inertia (11) or use (XXI). 

Illustrative Example. Find the moment of 
inertia for any gravity axis of a rectangle. 

Solution. Taking OX and OY as in the figure, 
then T _ Aa 2 T _ AW- 

and •••^ = f» »V 2 = f' 

and the equation of the ellipse of inertia is 



(1) 



= 3. 




MOMENTS OF MASS AND INERTIA 41 

The radius of gyration for any gravity axis is then the reciprocal of the radius 
vector of its point of intersection with the ellipse, or also, by (XXI), 
n = \{d 2 cos 2 6 + b 2 sin 2 d). 

PROBLEMS 

1. Show that the ellipse of inertia for any regular polygon is a circle. What is 
the conclusion regarding the moment of inertia with respect to any gravity axis ? 

2. Find /for a rectangle whose sides are 2 a and 2 b with respect to a diagonal. 

Ans. %M-^ 



3 a 2 + b 2 

3. Find I for an isosceles triangle with respect to an axis through its center of 
area and inclined at an angle a to its axis of symmetry, a being its altitude and 2 b 
itsDase - Ans. iiV/(ia 2 cos 2 « + 6 2 sin 2 a). 

4. Find I for an ellipse with respect to a diameter making an angle a with the 
major axis. 

Ans. 1 = -M(l) 2 cos 2 a + a 2 sin 2 a) = - M — , where r = - diameter. 
4 4 r 2 2 



CHAPTER II 

KINEMATICS OF A POINT. RECTILINEAR MOTION 

That portion of mechanics which is concerned with the study 
of motion is called dynamics. The subject matter of dynamics is 
divided into two parts, kinematics and kinetics. Kinematics treats 
of pure motion, that is, motion without reference to the mass of 
the body which is moving or the forces producing the motion. It 
has to do solely with the relations of time and space. Kinetics 
treats of motion, including consideration of the mass of the body 
moved and the forces acting upon it. This chapter treats of the 
kinematics of a point which moves on a straight line. 

24. Motion on a straight line. In order to indicate the posi- 
tion of a point upon a line, we select on that line a fixed point 0, 



called the origin. The position of any point P with respect to 
may then be determined by the length OP and its direction from 
the origin. For the application of mathematical analysis to the 
rectilinear motion of a point, it is necessary to regard the path as a 
directed line;* that is, we must assume an origin, a unit of length, 
and a direction. If the measure of the length OP be denoted by #, 
then it is obvious that x is variable if P is a moving point. The 
motion of P is said to be completely determined when the posi- 
tion of P is known at every instant of time ; that is, when the 
variable x is a function f of the time t, since the position is 
determined by the value of x. Hence for rectilinear motion we 
have the relation 

(I) * = <)>(*). 

This equation is called the equation of motion. Its significance 
is this, that from it we may find the position of the moving point 
at any instant of time. 

* Analytic Geometry, p. 23. f Calculus, p. 12. 

42 



KINEMATICS OF A POINT. RECTILINEAR MOTION 43 

In order to indicate instants of time it is necessary to select 
some fixed instant from which the time may be reckoned, forward 
and backward. 

This fixed instant, called the origin of time, is denoted by 
t = 0, and time before is indicated by a minus sign, time after by a 
plus sign. 

The position of the moving point when t == is called the 
initial position. The corresponding value of x is called the initial 
value of x and is denoted by x . From (I) we have, 

a* = 0(0). 

For example, if the equation of motion of a mov- 
ing point is x = t 2 — 2 £, we find the table of values of 
t and x as given. We see, therefore, that the point 

X—lx=0 x-3 x=8 . 



&t*2 



S=4 



2L 



t 


X 








1 


-1 


2 





3 


3 


4 


8 


etc. 


etc. 



moves from the initial position to the left and, 
after reaching the extreme position x = — 1, there- 
after moves continuously to the right. 

As a second example, consider the motion defined 
by the equation x — acos^Trt. Remembering from trigonometry 
that the cosine of a variable, increasing angle varies 
from 1 to — 1 inclusive, it is plain that with increas- 
ing time, x varies from a to — a inclusive ; that is, 



t 


X 





a 


1 





2 


— a 


3 





4 


a 


etc. 


etc. 



A' U AX 

the moving point P oscillates between the points 
A and A! of the figure. The initial position is A, 
since x = a, and the point is again at A after the 
lapse of four seconds. 
The vibratory motion just discussed is an example of simple 
harmonic motion, and will appear frequently in these pages. 

25. Velocity. By the velocity at any instant of a point in 
rectilinear motion is meant the time-rate of change of its position 
at that instant. When the equation of motion is x = cf> (£), the 
velocity is the rate of change of x with respect to t ; that is,* the 



Calculus, p. 148. 



44 THEORETICAL MECHANICS 

derivative of x with respect to t. Hence, denoting the velocity at 
any instant by v, we have 

(ii) v= if = cf) ' (f) - 

The value of the velocity at the origin of time is called the initial 
velocity, and is denoted by v . From (II), we have 

*>o=</>' (0). 

Dimensions. Velocity is defined as the limit of the quotient 
of distance Ax by time A*. The derived unit of velocity is 
therefore expressed in terms of the fundamental units of length 
and of time by the dimensional equation 

Velocity = — s — -. 
time 

If v = <f> r (t) is positive for the value t = t v we know that at 
the instant t = t v x = cj> (t) is an increasing function* of £, and the 
point is moving towards the right along the directed line OX. 
If v is negative, x is a decreasing function of t, and the point is 
moving towards the left. If for t = t v v 1 = <$>' (t^) = 0, the point 
at the instant t = t 1 is at rest. If the velocity is constant, the 
motion is said to be uniform. The numerical value of the velocity 
is called the speed. 

For example, to discuss the velocity of a point when its 
equation of motion is x = t 2 — 2 1, we find, by differentiation, 
v = 2 t — 2. Giving t successive values, we ob- 
tain the values in the table. The point is the 

-1 , 12 3 

' o 1 : = x 

initial position, and — 2 the initial velocity. 
The point is therefore moving in the negative 
direction along the line OX with a speed of 2 
units of distance per unit of time.f At the in- 
stant t = 1 the velocity is zero and the point 
is at rest. For values of t greater than 1 the velocity is positive, 
and the point moves in the positive direction along OX. 

* Calculus, p. 116. 

f That is, two feet per second, if the unit of distance is one foot and the unit of time 
one second. 



t 


X 


V 








-2 


1 


- 1 





2 





2 


3 


3 


4 


etc. 


etc. 


etc. 



KINEMATICS OF A POINT. RECTILINEAR MOTION 45 

26. Acceleration. If the velocity of a moving point is variable, 
the point is said to have acceleration. In mathematical terms, 
acceleration is the time-rate of change of velocity. That is, 
acceleration at any instant is the derivative of velocity with 
respect to the time. Hence, denoting acceleration at any instant 
by/, we have 



fjjjA ~_ dv _ d fdx\ _ d 2 x m 

^ J T ~ dt~ dt\dt)~ dV ' 



or, acceleration is the second derivative of the distance with 
respect to the time. When the equation of motion is x = <K0> 
we obtain, by differentiation, 

• /-£ =*"«>• 

The acceleration may be expressed in another form, which is 
frequently useful in the solution of problems in mechanics. We 
have x — $(£), and this may be solved for £, giving 

(1) t = ^{x). 

The velocity is a function of t ; namely, v = $ (t). When the 
value of t from (1) is substituted in this expression for the velocity, 
we have v expressed as a function of x. 

(2) v = F(x). 

This expression determines the velocity when the position is 
known. We have, from calculus,* 

dv __ dv dx 
dt dx dt 

fi'Y* ft 1) 

Since v = — , therefore, f=v — . 

dt dx 

For convenience, the preceding results may be summarized : 

I. Equation of motion,! as = <|> (*). 

II. Velocity at any instant, v = ^ = <(>' (*). 

TT , . *—dv_d 2 <c_M ur ^_„dv 

III. Acceleration at any instant, ' ~~dt~ ~d& dx 

* p. 57. 

t Other letters, e.g. y, s, will be used also to denote the position of the point P. 



46 THEOEETICAL MECHANICS 

The physical meaning of the algebraic sign of the acceleration 
is made apparent by the following consideration. If the point P 
moves along OX towards the right, the velocity is positive ; if 
towards the left, the velocity is negative. The acceleration is 
positive if v increases algebraically, and negative if v decreases 
algebraically. Hence, if 

P moves to the right with increasing speed, v > 0, f > 

P " " " " " decreasing " v > 0, / < 

P " " " left " increasing " v < 0, / < 

P " " " " " decreasing " v < 0, / > 0. 

If the acceleration is constant, the motion is said to be uni- 
formly accelerated. The special case when the acceleration is 
zero, and hence the velocity constant, has been already referred to 
in Art. 25 as that of uniform motion. 

Dimensions. Acceleration is denned as the limit of the quo- 
tient of velocity Av by time At. Its dimensions are therefore 
velocity divided by time, or distance divided by the square of the 
time. The relation between the derived unit of acceleration and 
the fundamental units of length and of time is expressed by the 
dimensional equation 

Acceleration = — &— -• 
time 2 

Two systems of units are in common use, the English and 
French. These are given in the table : 



Units 


English 








French 


distance 


foot 








centimeter 


time 


second 








second 


velocity 


1 ft. per sec. 








1 cm. per sec. 


acceleration 


1 ft. per sec. 


in 


1 


sec. 


1 cm. per sec. in 1 sec. 



27. Distance-time diagram. Discussion. The preceding dis- 
cussion has shown that distance, velocity, and acceleration of a 
moving point are functions of the time. The determination of 
the variation of these variables with the time constitutes the dis- 
cussion of the motion. The graph of the equation of motion is 
very useful in making the discussion. Since x is a function of t, 



KINEMATICS OF A POINT. BECTILINEAB, MOTION 47 

we may plot the curve represented by the equation x = $ (t), 

where t is the abscissa and x the ordinate. This curve is called 

the distance-time diagram. For a 

given instant of time, t v we have 

a given value of the abscissa ; e.g. y^ , 

OA in the figure. The correspond- -/- ■ 

ing value of x is the ordinate AB and 
the position* on the path OX is P v 

Since the velocity is the derivative of x with respect to t, its 
value is given geometrically by the slope of the tangent at B ; that 
is, by tan a. The numerical value of the acceleration is not given 
directly by the figure, but its sign is determined by noticing the 
form of the curve. If the curve is concave upwards, the sign of 
the acceleration is positive; if concave downwards, the sign is 
negative.f Maximum and minimum points on the graph of the 
equation of motion indicate extreme J positions of the point mov- 
ing on the straight line ; that is, positions where the velocity 
is zero. At such a point the velocity changes sign. With refer- 
ence to the moving point P this means that it ceases to move in 
one direction and begins to move in the opposite direction. A 
maximum point corresponds to an extreme position upwards, since 
the first derivative changes from plus to minus. For a maximum 
point the second derivative is negative ; hence for an extreme up- 
ward position the acceleration is negative. Similarly, a minimum 
point corresponds to an extreme downward position and the accel- 
eration is positive. A point of inflection on the graph of the equa- 
tion of motion indicates that at the corresponding instant of time 
the acceleration (which is the second derivative of x with respect 
to f) is zero. When the characteristics of the motion have been 
ascertained from this discussion, it will be convenient to take the 
path along a horizontal line. The properties already known of 
the motion on the X-axis are readily interpreted on the horizontal 
path. 

* The student must be careful not to confuse the distance-time curve with the path 
of the point. The path lies on OX, and the position of the point at any instant, t lf is 
found by constructing the point B in the diagram whose abscissa equals t\, and then 
projecting this point on to the distance axis, as Pi in the figure. 

I Calculus, Chapter IX. 

t The word "extreme " here means relative extreme, just as in geometry the word 
"maximum" means relative maximum. 



48 



THEORETICAL MECHANICS 



ILLUSTRATIVE EXAMPLES 

1. Discuss and draw the distance-time diagram for the motion denned by 

(1) x=t s -3t 2 + 2t. 
Solution. By differentiation, we obtain 

(2) v = 3 t 2 - 6 t + 2, 

(3) f=6t-6. 

The extreme positions of the moving point, and consequently the maximum and 
minimum points on the graph, are given by the condition, 

v = 3 t 2 - 6 t + 2 = 0, 
V3 



whence 



:1 ± 



or approximately 



ti = 0.4, h = 1.6. 
The corresponding values of x are approximately 

£i = 0.38, £ 2 = -0.38. 
For t < 0.4 the velocity is positive. 
For 0.4 < t < 1.6 the velocity is negative. 
For t > 1.6 the velocity is positive. 
The acceleration is zero, and consequently there is 
a point of inflection on the graph when t = 1. The cor- 
responding value of x is 0. For t < 1 the acceleration is 

negative, and since /=— , the velocity is decreasing 

(algebraically). For £>1, the acceleration is positive and the velocity is in- 
creasing. The distance-time diagram may now be drawn. We may summarize 
the results obtained in the following table : 




t 


X 


V 


/ 


0. 


0. 


2. 


-6. 


0.4 


0.38 


0. 


-3.6 


1. 


0. 


-1. 


0. 


1.6 


-0.38 


0. 


+ 3.6 


2. 


0. 


2. 


6. 


3. 


6. 


11. 


12. 


increases 


increases 


increases 


increases 



This table and the preceding graph are equivalent. From either we may 
make the discussion of the motion, and in the solution of problems each should 
serve as an aid to and a check upon the other. The discussion of the motion on 
a horizontal line follows. When t is zero the point P is at 0. As t increases 
from to 0.4, P moves to the right with a velocity which is decreasing numeri- 
cally. When t = 0.4, the velocity is zero and the point P is at an extreme position 



KINEMATICS OF A POINT. RECTILINEAR MOTION 49 

x = 0.38 to the right (the acceleration is negative). As t increases from 0.4 to 1.6, 
the point moves to the left. When t = 1.6, the velocity is again zero and the point 

-38 .38 6 Values ofx 

1.P20.4 3 Values oft 

is at an extreme position x = — 0.38 to the left (the acceleration is positive). As t 
increases from the value 1.6, the point moves always to the right with increasing 
velocity and acceleration. 

2. Discuss and draw the distance-time diagram for the motion defined by 

(1) x— a cos Jet. 
Solution. Differentiating, we obtain 

(2) v = — ak sin kt, 
and for the acceleration, differentiating (2), 

(3) / = - ak 2 cos kt = - k 2 x [from ( 1) ] . 

Hence the acceleration and distance are proportional and differ in sign. Such 
a motion is called a simple harmonic motion. 

The locus of (1) is a cosine curve, the properties of which are well known. 
The graph of the equation of motion has maxima when kt = 2 nir (n any integer), 

X 




minima when kt = (2n + l)'7r, and points of inflection when kt = ^ n — ^-ir. At 

any maximum point the ordinate is equal to a, and at any minimum point it is 
equal to — a. The variation of x, -y, and / is exhibited in the table. 



Angle kt 


t 


X 


V 


/' 








a 





-ak 2 


7T 
2 


7T 

2k 





— ak 





7T 


IT 

k 


— a 





ak 2 


3tt 
2 


3tt 

2k 





ak 





2tt 


2tt 
IT 


a 





-ak 2 


etc. 


etc. 


etc. 


etc. 


etc. 



It is now easy to make the discussion of the motion. "When t = 0, xo = a, v n = 0, 
f is negative, and the point starts from an extreme position to the right. As t 



50 THEORETICAL MECHANICS 

increases from to-, the figure [see also (2)1 shows that the slope of the tangent 
k 

line (and consequently also the velocity) is negative, and the point moves towards 

the left. When t = -, x = — a, v = 0, />0, and the point is at an extreme posi- 
k 

tion to the left. As t increases from _ to — , the velocity is positive and the point 

k k 

moves to the right. When t = — , we have again the initial values of x, v, and /. As 

fc 

t increases from the value — , the motion just described is repeated again and again. 
The motion is a vibration or oscillation between the points N and N' of the figure. 

y -a | a N Valu es ofx 

tJm ^Jjt Values oft 

«&- etc. 

The distance a is called the amplitude of the vibration. The time required to 

2 IT 

move from N to N 1 and back to N again is — . This is called the period of the 

k 
vibration. The point midway between JSF and N' (the point in the figure) is 
called the center of the vibration. 

The periodicity of the motion may be best established by reasoning thus. We 
note first that the series of values of any trigonometric function is repeated when the 
angle has increased 2 -k radians. Since x, v, and /are in this case dependent in their 
variation upon sine or cosine, Xhen it is plain that they assume their original values 
when kt has increased to kt + 2 it. But 

Hence t has changed to t -\ — — , and the increment — is accordingly the period, 
k k 

3. Discuss the motion defined by 

(1). x =Acos(kt + B). 

Solution. The distance-time diagram is again a cosine curve with A for maxi- 
mum displacement. The difference from the preceding example consists in this : 
the initial position on the path is not at an extreme position, but at x = A cos B. 
The conclusion is, therefore : 

The equation (1) represents a harmonic motion with the period — —, and this is 
true for all values of A and B. 

Equation (1) is the general solution of the equation 71, Chapter XIV, which is 
called the differential equation of harmonic motion. The statement just made ex- 
plains the designation. 

4. Discuss the motion defined by 

(1) x — ae~ f . 
Solution. Differentiating and using (1), we obtain 

(2) v = — ae~ l — — x, 

(3) / = ae-* = x. 



KINEMATICS OF A POINT. KECTILINEAK MOTION 51 

In this case, therefore, the acceleration and distance are proportional and agree 

in sign. From (1) x = — , and therefore x. which is always positive, decreases 

e* 
numerically as t increases ; v is always negative and 
decreases numerically (that is, the speed decreases). 
The graph is now readily drawn and exhibits 
the motion of a point from N towards O with 
constantly diminishing speed and acceleration. 
The motion dies away as is approached. There is obviously no period, and the 
motion is called aperiodic. 

5. Discuss and draw the graph of the equation of motion, 



X 


el 




a N 


a 


~o*~ 


X 









T 



(1) 



aer 



cos kt, 



Cosine Curve 



a, a, k being arbitrary, positive constants. 

Solution. Differentiating, we obtain for v and /the expressions, 

(2) v = - ae~ a \a cos kt + k sin kt) , 

(3) /= ae- at [2aksmkt + (a 2 - k 2 ) cos kt]. 

The graph of (1) is readily constructed and the characteristics of the motion ap- 
pear from it. Write (1) in the form of a product, 

(4) x = ae~ at • cos kt. 

The factor cos kt varies 
from — 1 to + 1. Hence the 
distance x varies from — ae~ at 
to + ae~ at ; that is, the graph 
of (4) is bounded by the curves 

(5) x = — ae~ at , x = ae~ at . 

These are the dotted lines 
of the figure. 

Again, the product in (4) 
vanishes only when one of the 
factors is zero. But e~ at is 
never zero for finite time. Hence x = when and only when cos kt = 0. 

Furthermore, the graph touches * the boundary curve when cos kt = ± 1. We 
therefore draw also the auxiliary curve X\ = cos kt. We now observe that 

(1) The points of contact with the boundary curves are directly over (or under) 
the maximum and minimum points on the cosine curve. 

(2) The required curve crosses the T-axis at the same points as the cosine curve. 
The graph may now be drawn, for we have merely to construct a winding 

curve from the initial point t — 0, x = a, which shall cross OT at M\, Ms, M$, etc., 
and touch the boundary curves at points corresponding to Jf 2 , Mi, etc. 

From this construction it is obvious that maximum and minimum values of X 
occur between each intersection on OX and the succeeding point of contact with 




* For when cos kt = ± 1, then sin kt — 0, and we find from (2) v = =F aae~ at . This 

dx 
equals—^> from (5). Hence the slope of (4) at M 2 , M±, etc., is the slope of the proper 

boundary curve (5). 



52 THEORETICAL MECHANICS 

the boundary curve ; that is, for a value of t between successive odd and even 

multiples of — • In fact, from (2), v — 0, when 
2i k 

(6) a cos kt + k sin kt = or tan kt — • 

k 

Now the tangent is negative when the angle is of the second or fourth quad- 
rants. Hence kt must lie between — and — , or — and — , etc. , or t is between 

2 2 2 2' 

successive odd and even multiples of — • 

2 k 

The characteristics of the motion are now obvious. It may be described as a 
vibration with constantly diminishing amplitude. Kemembering that the simple 
harmonic motion represented by the factor a cos kt of (1) has the constant ampli- 
tude a, it is plain that the presence of the second factor e~ at accounts for the dimin- 
ishing amplitude. 

This factor diminishes as t increases, and is called the damping factor. The 
motion is called damped vibration. 

From (6), it appears that a period of time equal to — (from kt = 2ir) must 

elapse between successive maxima. The motion is accordingly said to have a period 

2 IT 

equal to — , the same, namely, as the period of the undamped harmonic vibration 

k 
(Ex. 2). 

The successive amplitudes obey a simple law. For such positions differ by a 
semi-period, and hence two such values of x may be written in the form 

xi = ae~ at , x 2 = ae- a ( t+ f)' 
Taking natural logarithms and subtracting, we obtain 

\0gXi-\0gX2=^- 

k 

That is, the logarithms of successive amplitudes form a decreasing arithmetical 
progression. 

This is otherwise expressed by the statement that the logarithmic decrement of 
the amplitude is constant. 

6. Discuss the motion whose equation is 

(1) x— Ce-M* cos (It + 7), 

in which C, /*, I, and y are arbitrary constants, ix being positive. 

Solution. The construction of the graph is precisely as in the previous ex- 
ample ; namely, the boundary curves are 

(2) x = ± Ce-n f , 
and the auxiliary cosine curve is 

(3) x 1 =cos (U + y). 

The difference from the preceding case is in the initial position, which is now 
x = C cos 7, an arbitrary point on the path, not necessarily (7 =£ 0) an extreme 
position. 

The result is then this : 

The motion defined by (1) is a damped vibration with the period — ., and this is 
true for all values of C, 7, Z, and /*, provided /x > 0. 



KINEMATICS OF A POINT. RECTILINEAR MOTION 53 

Equation (1) has the form of the general solution of the equation 73, Chapter XIV, 
which is called the differential equation of damped vibration. The theorem just 
stated explains this designation. 



PROBLEMS 

1. Show that each of the following motions is uniform or uniformly accel- 
erated, draw the distance-time diagrams and discuss the motion : 

(a) x = 2 — 4 t ; (h) s = v t + h ; 

(&) y = at+b; (i) s = \ gt 2 + v b t + s ; 

(c) s = 6t- 16 t 2 ; ( j) y = 50 + 10 t - 16 f 2 ; 

(<-?) y - 10 — * - 3 « 2 ; (*)'s = ^ sin a •; t 2 ; 

(e) x = a + bt + cf 2 ; (?) s = v £ — \ g sin a • t 2 ', 

(/) s = i f/£ 2 + v £ ; (wi) x = 1000 * - 16 t 2 ; 

(g) y = v Q t-l gt 2 ; (>*) y = - 1000 * + 10 ? 2 . 

2. Show that the distance-time diagrams of uniform and of uniformly acceler- 
ated motion are respectively a straight line and a parabola. 

3. Show that each of the following is a simple harmonic motion.* Draw the 
distance-time diagrams, discuss the motion, and find the amplitude a and period T 
in each case. 

(a) x = 5 sin t ; (g) y — 10 sin (£ wt — | ir) ; 

(&)?/ = 10 cos £ ; (ft) ?/ = sin £ + cos£; 

(c) s = 2 cos i irt ; (i) s = a sin (to + a) ; 

(d) x = 5 sin f ?r£ ; ( j) as = 6 cos (fit — /3) ; 

(e) y = a sin to ; (jfc) x = 2 sin £ + 3 cos t. 
(/) x = 5 cos («+ i?r) ; (?) a: = a cos to + 6 sin to. 

J.7isw>e>-s denoting amplitude by a and period by T. 
(a) a = 5, T= 2 tt ; (h) a = \/2~, T= 2 tt ; 

(&) a = 10, r=2x; ( . } i = a>r== 2r. 

(c) a = 2, T=4; & 

W « = 5, 7 = |; (j) fl = ^ r = 2 J r. 

« ■ = « r =F ! (*) « = Vl3,r=2.; 

(/) « = 5, r = 27r; 2 

(,,) a = 10, r=4; d) «=Vtf + *, r=^. 

* Show that the given equation is obtained from x = A cos (to + B) by replacing the 
constants A, Jc, B by particular values. Thus for (a), a; = 5 sin t, we set A = 5, k = 1, 



For x = 5 cos (t — ^) = 5 cos (| — A = 5 sin *. 



54 THEORETICAL MECHANICS 

4. Show that the acceleration and the distance are proportional and differ in 
sign for each of the following motions (simple harmonic) : 

(a) x = A sin (kt + a) + B cos kt ; 

(b) y = A sin kt + a cos (kt + /3) ; 

(c) s = a sin (fit — a) + b cos (fit — /3). 
Reduce each to the form A cos (kt + 5) . 

5. Discuss and draw the distance-time diagram of each of the following motions 
and show that each is a damped vibration.* 

(a) x = 5 e~* t cos \ irt ; (g) x = ae~^ sin kt ; 

(b) y = 2e~^ t smlTrf ) (h) y - 5e~^sm(t + \ tt) ; 

(c) s = 10e~^cosc; (0 s = e~ at (a sin kt + b cos kt) ; 



(d) x = 5 e -2 * sins; 



(j) x— 10 e ^ 3 cos £ ; 



r = e 'cos(* + ^), s = e-^ S m(^ + ^); 



(/)y = 8 ^*8in {?(«-+!)}; CD 2/ = e a 'cos>pt + /3). 



6. Discuss and draw t the distance-time diagrams of the following equations of 
motion: 

(a) x = sin t + cos 2 £ ; (d) V = e~* * cos c + sin £ ; 

(6) x = alog(l — £); (e) y = sini £ _|_ sin-^ ; 

(c) 2/ = Ke* + e~*); (/) y ■=«-*' cos « + 10 sin *. 

L /72o 

7. Show that every solution of h v-s = X, where /* and X are constants and 

dt 2 
/*<0, defines a harmonic motion. Find the period and the center. 

Ans. T=^; ( \ oY 

J2o r 7<; 

8. When will solutions of— + 2/*— + Xs = define damped vibrations ? 

dt 2 dt 

Arts. X > fi 2 . 

9. Discuss and draw the distance-time diagrams of the following equations of 
motion : 

(a) x=tsint; (c) s = (t + 1) cost ; (e) y =™£l- t 

6 + 1 



(b) y — e f cos t ; 



(rf) x = ^BJ; (/) s = tsmft + ^\ 



10. Discuss and draw the distance-time diagrams of the following equations of 
motion : 

(a) x — sin t + 2 ; (e) x — ae~ at cos &£ + b ; 

(b) y = cost — 10; (/) x = a cos kt + & ; 

(c) s = e - ' cos t + 1 ; (g) y = a cos &c + 6 sin kt + c ; 

(d) z = 10e"^ cos t + 5 ; (A) s = A sin (Atf + (3) + 6. 

* Show that the given equation is obtained from (1), p. 52, by giving to C, m, I, and 
V particular values. 

t When the function of the time is the sum of two simple functions, we may draw 
the graphs of the latter and add the corresponding ordinates. For example, in (a), add 
the ordinates of x-y ■— sin t and x 2 = cos 2 1. 



CHAPTER III 

KINEMATICS OF A POINT. CURVILINEAR MOTION 

28. Position in a plane or in space. Vectors. In the discus- 
sion of the rectilinear motion of a point the quantities involved 
were time t, position on the straight line x, velocity v, speed «, 
and acceleration /. Any value of t, x, v, or / is indicated by a 
single number (positive or zero or negative), and any value of s is 
indicated by a single number (positive or zero). Quantities which 
take on values that can be indicated by single numbers are called 
scalar quantities. Such quantities have magnitude (-f- or — ) only. 
A vector quantity is one which has magnitude and direction. For 
example, (1) the position of a point P(/>, 0) in a plane is indicated 
by its distance from the origin (magnitude) and the angle which 
OP makes with the initial line; (2) the position of a point PQ>, (/>, 0) 
in space is indicated by its distance from the origin and the 
direction of the line OP.* Since a scalar quantity has magnitude 
only, any value which it may take on can be represented graphic- 
ally by the length of a line taken in the proper algebraic sense. 
To represent a vector quantity graphically the line must have 
length and direction. By indicating the direction properly the 
length may always be taken as positive. Hence we make the 
definition, a vector is a straight line having length and direction. 
From this definition we conclude that two vectors 
AB and A"B" are equal if the lines AB and A"B" 
are parallel, equal in length, and taken in the same 
sense. If the lines are parallel and equal in length, 
but taken in the opposite sense, that is, if the di- 
rections differ by 180°, as AB and A'B', we say AB = - A'B'. 
A vector is zero if, and only if, its length is zero. In solving 
problems involving vectors we may always replace a vector by 
an equal vector, which is equivalent to saying that a vector may 
be moved providing it is kept always parallel to its original posi- 
tion. 

* Analytic Geometry, p. 394. 
55 





56 THEOKETICAL MECHANICS 

29. Addition of two vectors. ■ If a point is moved in a plane, 
the displacement is a vector quantity. Suppose a point is moved 
from the origin to the position A(2, 4). The displacement is rep- 
resented by a vector whose length A = V20 and whose direc- 
tion is indicated by the angle which 
the line OA makes with the X-axis. 
Suppose the point is given a second 
displacement from A (2, 4) to B (5, 3). 
"^JC This displacement is represented by 
the vector AB, whose magnitude is V(2 — 5) 2 + (4 — 3) 2 = Vl0, 
and whose direction is given by the angle </>. These two dis- 
placements taken in order are evidently equivalent to a single 
displacement from to B, which is represented by the vector OB, 
the magnitude of which is V34 and whose direction is given by 
the angle i/r. Hence we say that the vector OB is the sum of the 
vectors OA and AB. 

OB=OA + AB. 

If two vectors AB and BE are given, we obtain the sum 
AB + DE \\i the following manner. From the point B construct 
a vector BO=DE. The vector AC is now 
denned as the sum of AB and BO, and, there- 
fore, as the sum of AB and BE. 

AB + BC=AC. 
.-. AB + BE = AC. 

The process of adding two vectors is essentially this. Bring 
the two vectors into such a position that they form a broken line 
ABC. Their sum is then equal to the closing 
* line AC. It is readily seen that the order of 
addition can be changed without altering the 
sum. 

AB + BC= BC+ AB. 

The figure is a parallelogram and the proof is obvious. 

Addition of any number of vectors. The preceding process is 
applicable to the addition of any number of vectors. Suppose it 
is required to find the sum of the vectors A X B V B 2 O v C 2 B V and 
B 2 E V This is accomplished by repeated application of the process 
of adding two vectors. 






KINEMATICS OF A POINT. CURVILINEAR MOTION 57 

(1) Construct the vectors AB and BC equal respectively to 
A 1 B 1 and B 2 C V The sum of these two vectors is AC. 

.-. A 1 B 1 + B 2 C 1 = AC. 

(2) Construct OB = C 2 B V The sum 
of AC and CD is AB. That is, 

A 1 B 1 + B< L C 1 + C 2 B X = AB. 

(3) Construct BE=B 2 E V The sum 
of AB and BE is AE. Therefore, 

A l B \ + B Z°1 + ^2 A + -JV^i = -^ 

The process is applicable to any num- 
ber of vectors and is essentially this. To add any number of 
vectors, form a broken line having its segments equal, respec- 
tively, to the given vectors; the sum is then the closing line.* 
Since the order of addition of two vectors may be changed with- 
out changing the sum, the order of addition of any number of 
vectors may be changed without changing the sum. 

The sum of any number of vectors is called the resultant of 
those vectors. 

30. Subtraction of vectors. Any vector AB may be sub- 
tracted from the vector CB by adding to CB the negative of AB. 
In the figure BE=-AB and CE = CB + BE = CB - AB. 
For practical purposes it is more convenient to obtain the 
difference of two vectors as follows : To 
subtract AB from CB, lay off the two vectors 
from the same origin ; that is, construct 
CE = AB. Then 

CF+ EB = AB + FB = CB. 

Whence, by transposing the term AB, 
EB = CB - AB. 



*~B 




Fig.b 



The results of the two methods are equal, as can be shown by 
comparing the equal triangles, figure a and figure b. 

31. Multiplication of a vector by a scalar. If a vector AB is 
multiplied by a positive scalar W, the result is a vector A'B' 

* Analytic Geometry, p. 47. 



58 THEOEETICAL MECHANICS 

having the same direction as AB, while the magnitude of A r B f is 
W times the magnitude of AB. For example, in the figure (a), 

A'B'= 2AB. 

A ^ B If a vector AB is multiplied by a nega- 

' , ^ , tive scalar — IF, the result is a vector 2?'.A' 

Fi 9- a which has a direction opposite to that of 

AB and a magnitude equal to W times the 
, magnitude *of AB. For example, in the 
Fig.b figure (5), 

B ! A' = -2AB. 

To divide a vector by a scalar W, we multiply the vector by— 
For example, in figure a, 

AB = \A!B } 

and in figure b, AB = - \ B'A'. 

32. Resolution of plane vectors. Suppose a vector AB is given 
and it is required to find two vectors which are equivalent to AB, 
that is, whose sum is equal to AB. This may be done in an infi- 
nite number of ways. For, suppose is any point, and the lines 
^4. (7 and OB are drawn. Then, by the definition of a vector sum, 

AC+ CB=AB. 

The point may be determined so that the vectors A and OB 
are parallel to the X- and Y-axes respectively. This is accom- 
plished by drawing through A a line parallel to the X-axis and 
through B a line parallel to the Y-axis. These two lines inter- 
sect in the required point 0. For convenience we will denote the 
vector AB by a, the vector AO by * a x , and OB by a y . The vector 
a x is called the component of a in the direction 
of the X-axis and the vector a is said to be 
resolved along the line OX; a y is the com- 
ponent of a in the direction of the I 7 -axis, and 
a is said to be resolved along the line OY. It 
is evident that a x is the projection of a on the 
X-axis and a y is the projection of a on the Y-axis. 

A vector may be resolved along any directed line by projecting 
the vector on that line ; that is, the component of a vector along 

*In using the components of a vector a, we need to give only the numerical values. 
The directions are indicated by the subscripts. 



2^ 



KINEMATICS OF A POINT. CURVILINEAR MOTION 59 

any directed line equals its magnitude multiplied by the cosine of 
the angle its direction makes with the given line. 

In solving problems involving vectors it Y 
is usually more convenient to deal with the 
components. If the axial components of a 
vector a are a x and a y , it is evident from the 
figure that the magnitude* of a is given by 
a = Vfl^ 2 -f- a y 2 and the direction of a is the 
same as the direction of a line from the origin to the point 
(ajfl ay). If we denote the angle which the vector a makes with 
the X-axis by (a?, a), we have 

cos (as, a) = — ; sin fa?, a) = — • 
v y a v a 

Hence we have the formulas : 




(I) 



a = +Va x 2 + a y 2 , 

a x — a cos (a?, a), 

a = a sin (as, a) = a cos (y, a). 



In particular the components of the vector which represents 
the position of a point P in a plane are the rectangular coordi- 
nates of that point. 

When the axial components a^ a y of a vector a are known, its 
component in the direction of any line I is 
readily found. Denote the angle which I 
makes with the X-axis by (a?, I). The projec- 
tion a t of a upon I is equal to the sum of the 
projections of a x and a y upon I. The projec- 
tion of a x upon I is a x cos (oc, t) and the pro- 
jection of a y upon I is a y sin (as, £). There- 
fore the component of a in the direction I is 




en) 



a l = a x cos (a?, Z) + «„ sin (as, £), 



Components of the resultant of any number of plane vectors. 
Let a, b, c • • • be given vectors with components a x , a y ; b x , b y ; 
c x , c y ••• respectively. Let R (components i? x , R tJ ) be the 
resultant of a, b, c •••. By the definition of a vector sum, we 
regard a, b, c ••• as the segments of a broken line, while R is the 



* The letter a represents the magnitude of the vector a. 



60 



THEORETICAL MECHANICS 



closing line. By the second theorem of projection,* the sum of 
the projections of a, b, c ••• upon any line is equal to the pro- 
jection of R upon that line. Therefore, 

«* + &* + c *+ *••> 
<*> v + h v + c v -\- -• 



(III) 



P 

to 



33. Vectors in space. The results for plane vectors may be 
extended at once to vectors in space. Any vector in space may 
be resolved along three mutually perpendicular lines by projecting 
the vector upon each of the lines. 

If the three mutually perpendicular lines are the JT-, JT-, and 
i^-axes, the components of the vector a are denoted by a^ a y , a z . 
The magnitude of a is a — Va x 2 + a y 2 + a z \ and its direction is 
the same as the direction of a line from the origin to the point 
(a x , a y , a z ). The direction cosines of the vector are 



cos (as, a) = — ; cos (y, a) : 
For space we have the formulas : 

(IV) 



cos (z, a) = 



= + V «/ + a v 2 + a*, 



a, 



a cos (as, a), 
a cos (y, a), 



a z = a cos O, a). 

Since the second theorem of projection holds also in space,f the 
components of the resultant R of any number of vectors a, b, c ••• 
are 

"-«* = «* + &* + «* + •-., 

(V) ^ .»„ = ?„ + &„+-<,„ + -, 

JR z = a z + b z + c 2 + — . 

When the axial components a x , a y , a z 
of a vector a in space are known, its com- 
ponent in the direction of any line I may 
be found. Let the direction angles of I 
be (as, f), (y, I), (s, I). The projection 
a t of a upon I is equal to the sum of the 
projections of a x , a y , and a z upon I. 
The projection of a x on I is a x cos (as, t). 

* Analytic Geometry, p. 47. t Analytic Geometry, p. 328. 




(3. 



KINEMATICS OF A POINT. CURVILINEAR MOTION 61 

The projection of a y on I is a y cos (y, I). 
The projection of a. on I is a e cos (s, £). 
Therefore, 
(VI) a t = a r cos (a>, I) + a y cos (2/, J) + a 2 cos (», 2). 

ILLUSTRATIVE EXAMPLES 

1. Find the resultant of the three plane vectors a, b, c whose components are 
— 2), (2, 6), (—7, — 1), respectively. 

Solution. 

^ fi2 x = 3 + 2-7=_2, 

B *Q IT >> 1^^-2 + 6-1 = 3. 



By (I) , B= Vl3, cos (a?, JB) = - 
sin (a?, jB) = 



Vl3 




Vl3 

In the figure, this result is checked by graphical construction of the resultant. 

2. Three vectors a, b, c in space have magnitudes equal to 12, 8, and 6, 
respectively, and their direction angles are as follows : 

(a) (a?, a) = f r, (1/, a) = } *-, (s, a) = £ *■ ; 
(6) (»,&)=**, (|f, ft) = fr, (2, &) = ±7r; 
(c) (a?, c) = i7r, (y.e)=lT, (*, C )=fx. 
Determine the resultant, 

Solution. Finding the axial components by (IV), we have 
«x=-6V2, a y = 0, «^ = 6V2, 
6 X = -4V3, 6, =-4, &,= 0, 
c x = 3V2, c y =— 4, c z = — 3. 
Hence, applying (V), the resultant has the components 
fi?, = -3V2- 4 V3, 

I Jk = 6V2- 3. 



3. Given in the XF-plane a vector a whose axial components are (-2, 1). 
Find its component along the directed line from the origin to the point (2, 1) . 

Solution. In the figure, I represents the given 
line, OA the given vector, OA' the projection of 
OA on I. 

2 1 

By geometry, cos (ac. I) = — - ; sin(a>, I) = — z . 

V5 V5 




By (II), 



i-L=. 



-§V5, 



V5 V5 V5 5 

The negative sign indicates that a x has the negative direction on I. 



62 THEOKETICAL MECHANICS 

PROBLEMS 

1. Determine in direction and magnitude the resultant of each of the following 
groups of vectors in a plane, given by their axial components. Verify the result in 
each case by a graphical construction. 

(a) (2,0), (-2,-6), (5,3). 
(5) (-1,5), (2, -1), (8,2). 

(c) (0,1), (5,6), (-2,-8), (-3,-4). 

(d) (9, 0), (10, 5), (6, 2), (1, 4), (2, 3). 
(«) (0, - 9), (-1,-6), (2, 5), (- 1, - 8). 

2. In the following examples the magnitude and angle made with OX of cer- 
tain vectors are given. Determine the resultant in each case. 

(a) 5, i7r ; 8, |7r. Ans. Components are [(f — 4V3), (|V3 + 4)]. 

(6) 2, Itt; 9, $x. Ans. (-|V3, - f). 

(c) 3, |tt; 1, |7r. Ans. (-— , — - lV 

V V2 y/2 / 

(d) 4, f 7T ; 10, 1 7T. ^»s. (- 5V3, 1). 

VV2 2 V2 2/ 

3. In problem 2 find the component of the first vector along the directed line 
determined by the second. 

Ans. («) §; (6) -1; (c) *L ; (d) -2; («) V2(V3 + 1). 

2 V2 

4. Given the axial components of the following vectors in space. Find the 
resultant of each group : 

(a) (1, 1, 5), (2, -1, 6). (c) (0, 6, 5), (1, 9, -8). 

(6) (1, 0, 8), (- 1, -1, 0). (d) (3, -4, 9), (6, 2, 3). 

(<0 (-1,2,8), (4,6, -2), (9,10,11). 

5. The magnitude and direction angles of certain vectors in space are as follows. 
Determine the resultant in direction and magnitude. 

(«) 10, i 7T, 1 1T, | 7T ; 5, 1 7T, 1 7T, f 7T. 

Ans. Components are -, — , 5( — ^— V3) • 

(6) 6, 1-7T, |7T, frj 4, iTT, ITT, |7T. 

^4?i.s. Components are [(3V2 + 2\/3), —3, —5]. 

6. Determine the component of each of the pair of vectors in problem 5 (a), 
(ft), along the other. 

7. A point has uniform motion along OX with a velocity of 10 ft. per second. 
Find the component of the velocity along the directed line from (0, 0) to (3, 4). 

Ans. 6 ft. per second. 

8. Find the resultant of the following velocities, the capital letters indicating 
points of the compass as usual, and the numbers the magnitude : 

15 N., 20 E., 20V2 N.W., 35 W. Ans. 35V2 N.W. 



KINEMATICS OF A POINT. CURVILINEAR MOTION 63 

9. Find the resultant of : 

(a) accelerations 5, 8, 10 parallel to the sides of an equilateral triangle taken 
in order. Ans. (Taking first component along X-axis) (— 4, — VS). 

(6) velocities 2, 5, 6, 3 parallel to the sides of a square taken in order. 

Ans. (-14,2). 

10. A point undergoes three displacements of 1, 2, and 3 units, respectively, in 
directions parallel to the sides of an equilateral triangle taken in order. What is 
the resulting displacement ? 

Ans. V3 in a direction perpendicular to the second side. 

11. A ship is carried by the wind 3 mi. due north, by the current 4 mi. due 
west, and by her screw 20 mi. southeast. What is her actual displacement ? 

12. A mail bag is thrown from a train with speed of 20 ft. per second perpen- 
dicular to the track. If the speed of the train is 40 mi. per hour, what is the 
direction and speed of the bag relative to the earth ? 

13. A particle is kept at rest by forces of 6, 8, 11 units. Find the angle be- 
tween the forces 6 and 8. Ans. 77° 21' 52". 

14. A boat is carried southwest by the current with a speed of 5 mi. per 
hour and 30° south of east by the wind at the rate of 12 mi. per hour. What must 
be the direction and magnitude of the speed due to her screw if she remains at rest ? 

15. Three posts are placed in the ground so as to form an equilateral triangle, 
and an elastric string is stretched around them, the tension of which is 6 lb. 
Find the pressure on each post. Ans. 6V3. 

16. ABCD is a square, and the middle point of BC is E. Find the resultant 
of three velocities represented by AB, AE, and AC. 

17. The angle between • two unknown forces is 62°, and their resultant divides 
this angle into 40° and 22°. Find the ratio of the component forces. 

18. Three forces act at a point and include angles of 90° and 45°. The first two 
forces are each equal to 2 units and the resultant of them all is VlO units. Find the 
third force. Ans. V2 units. 

19. If three forces of 99, 100, and 101 units, respectively, act on a point at angles 
of 120°, find the magnitude of their resultant and its inclination to the second force. 

Ans. V3, 90°. 

20. A weight of 40 lb. is suspended by two strings, inclined to the vertical 
at angles of 30° and 45°, respectively. Find the tension in each string. 

Ans. 20(V6 -V2), 40(V3- 1). 

21. Given the vectors a(3, - 2), b(5, 0), c(- 10, 6), d(7, 7). Construct the 
figures and find the resultants of the following : 

(a) a + 2 b - 3 c ; 
(6) 2a-b + c + 2d; 

(c) 3a + 4c-d; 

(d) 4b-2c + 5d; 

(e) 10 a + 5b + 4c ; 
(/) 2 a + 3 b + c - d ; 
(g) 2a-3b-2c-2d. 




64 THEORETICAL MECHANICS 

34. Displacement in a plane. Path. Suppose a point moves in 
a plane. Then its position vector changes as the time changes. 
If the law of the motion is known, the position vector p is a 
known function of the time, and its com- 
ponents (p x =x, p y = y) are known functions 
of the time ; that is, 

(VII) « = ♦(«)" v = +(*). 

Equations (VII) are called the equations 
of motion. By assuming values t v t 2 , £ 3 , etc., 
for the time we may compute the corre- 
sponding position vectors p r p 2 , p 3 , etc. The locus of the extremities 
of the position vectors is the path of the moving point P.. Since the 
components of- the position vector are the ordinary rectangular 
coordinates, the equations (VII) may be regarded as the parametric 
equations* of the path. The rectangular equation of the path 
may be obtained by eliminating t from the two equations (VII). 

If p x is the position vector at the instant t v and if p 2 is the 
position vector at the instant £ 2 , the total displacement during the 
interval of time from t x to t 2 is represented by the vector d = P\P^- 
This displacement is evidently equal to the difference of the vec- 
tors p 2 and p x (Art. 30); that is, 

d = p 2 - P r 

The position of the point at the instant £=0 is called the initial 
position. It is represented by the position vector p whose com- 
ponents are x = <j> (0), «/ = yjr (0). The length s of the arc de- 
scribed in the interval of time from to t is a function of t. The 
expression for s is given by (66, Chap. XIV) 

• = /[(ty + ®^-jc ,t ^® i, + } ^ j ** 

and .-. g = Vf^'(0! 2 +Sf'(OS 2 - 

The sign of the radical is always taken as positive. The deriva- 
tive — is the time-rate of change of s and is called the speed. 

* Calculus, p. 93. 



KINEMATICS OF A POINT. CURVILINEAR MOTION 65 




35. Velocity in the plane. Velocity curve. Suppose a point P 
moves along a path AB in the Jf 1^-plane and that the equations 
of its motion are x = ^^ y = ^(f). 

Suppose that at the instant t = t x the point is at the position P x 
represented by the position vector p r and at the instant t = t n it 
is at the position P 2 represented by the po- 
sition vector p 2 . During the interval of time 
t 2 — t x the displacement is represented by the 
vector d=p 2 — p r 

The quotient p — p 

is called the average velocity during the in- 
terval of time t 2 — t v The average velocity is a vector, since it 
is the quotient of a vector and a scalar. It has the same direc- 
tion as the displacement vector d = p 2 — p r and its magnitude is 
equal to the magnitude of d divided by t 2 — t v 

Let us now consider a fixed instant t = t v the corresponding 
position vector being j> v and denote an interval of time immedi- 
ately following t x by A£, and the displacement during the interval 
At by Ap. The average velocity during the interval of time At 

is therefore — ?• 

A* 

To fix the ideas, let us consider some particular values for At, 
the unit of time being 1 second. 

(1) Let At = l; the displacement 
vector Ap = -P 1 P 4 (see figure), and the 
average velocity during the interval of 
one second immediately following the 

P P 

instant t=t 1 is — ! — 4 - The vector rep- 
resenting the average velocity is there- 
fore equal to the displacement vector, 
that is, equal to the chord P X P^ 
(2) Let At = \ ; the displacement vector Ap = PiP 3 , and the 
average velocity during the interval of one half second immedi- 
ately following the instant t 




t± is 



1 

2 



9 p p 



P P ' 



66 THEORETICAL MECHANICS 

The vector representing the average velocity has the direction 
of the chord and its magnitude is equal to twice the length of the 
chord. 

(3) Let At = ^; the displacement vector Ap = P^P^ an( ^ ^ ne 
average velocity during the interval of one fourth of a second 
immediately following the instant t = t t is 

ZjZj — 4 P P — P P ' 

4 

The vector representing the average velocity has the direction 
of the chord P X P 2 and its magnitude is equal to four times the 
length of the chord. 

(4) Let At approach zero as a limit. The vector which repre- 
sents the average velocity — 2. has the same direction" as the chord, 
and hence when At approaches zero its direction approaches the 
direction of the tangent to the curve. Multiplying — £ by — ■ (As 

represents the increment of arc along the curve in the time Af), 
we may write : 

Magnitude of average velocity = — *- • — = — ■ '■ — ■£-• 

B s ^ At As At As 

As At approaches zero as a limit, As also approaches zero as a 

limit, and —*- = approaches 1 as a limit, while — ap- 

As arc FF At ^ 

proaches — as a limit. Therefore the magnitude of the average 

velocity approaches ~r. as a l imi t- 

We now make the definition : The velocity of the moving point at 
the instant t = t x is equal to the limit of the average velocity as At 
approaches zero. The magnitude* of the velocity is therefore 

and its direction is the direction of the tangent to the path. The 
cosine and sine of the inclination of the tangent are respectively:! 

dx dy 



* The magnitude of the velocity is the speed. The velocity is a vector quantity and 
possesses magnitude and direction, 
t Calculus, p. 142. 



KINEMATICS OF A POINT. CURVILINEAR MOTION 67 

Therefore, applying (I), we have for the axial components of 
the velocity : 



(VIII) 



ds doc doc jli,+\ 
Vx = dt cte = -dt = V^ 

y dt ds dt V K J 



From (VIII) we see that the component of the velocity of the mov- 
ing point P in the direction of the X-axis is obtained by differentiating 
the abscissa of P with respect to the time, and the component of the 
velocity in the direction of the Y-axis is obtained by differentiating 
the ordinate of P with respect to the time. In other words, v x is 
the velocity of the projection of the moving point on the .X-axis, 
and v y is the velocity of the projection of the moving point on the 
Z~-axis. Hence the 

Theorem. The axial components of the velocity in curvi- 
linear motion are equal to the velocities of the axial components 
of the motion. 

In general the velocity is different at different points of the 
path. At the point P 1 of the curve the velocity will be repre- 
sented by a vector v a ; at the point P 2 by a vector v 2 , etc. Let a 
new system of rectangular axes be chosen, 0, X, Y, and from the 
origin lay off the vectors v v v 2 , etc. 



r 


"V 5 

y 


\v 2 











X 




The locus of the extremities of the velocity vectors in the XY- 
plane is a curve which is called the velocity curve of the motion 
defined by (VII). The position vector of any point P of the velocity 
curve is equal to the velocity vector of the corresponding point 
P of the path. The rectangular coordinates of P(x, y) are equal 
respectively to v x and v y . 



68 



THE0EET1CAL MECHANICS 



Illustrative Example. Construct the path and the velocity curve for the 
plane motion denned by the equations 

(1) x = t,y = cos 2 t. 

Solution. Eliminating t, the equation 
of the path is found to be y = cos 2 x. 

Differentiating (1), the components of 
velocity are 

(2) v x = 1, v y =-2sm2t. 
Erom equations (2), it is seen that the 

velocity curve consists of a portion of the 
straight line x = 1. Since the sine cannot 
be numerically greater than 1, we have 
no points on the velocity curve for which y 
is numerically greater than 2. 



t 


X 


y 


v x 


Vy 





0. 


l 







7T 


7T 







-2 


4 


4 








IT 


7T 


-l 







2 


2 








3tt 


3tt 







2 


4 


4 








IT 


7T 


i 









Y 




(1,2) 







d,o)x 

(1-2) 



The coordinates of the moving point P and the components of velocity for cer- 
tain values of t are shown in the table. 



PROBLEMS 

1. Construct the path and the velocity curve for the plane motions defined by 
the following equations : 



(a) x = cos t , y = sin t ; 
(&) x = a cos t, y = 6 sin t ; 

(c) x = 2 sin t, y = cos 2 £ ; 

(d) x = cost, y = 2 sin J £ ; 

(e) x = a cos £, y — a cos 2 t ; 
(/) x = a sin 2 Z, y = a sin £ ; 

(g) x = a(t— s'mt), y=a(1— cost); 
(ft) x = a(£+sinf),?/=:a(l— costf); 
(i) a: = a(t— s\\\t),y = b (1— cos£); 
( j) x = a cos 3 t, y — a sin 3 £ ; 



(&) x = a cos 3 t, y — b sin 3 £ ; 
(J) x = 6 £ - £ 2 , y = 3 t ; 
(?») x = at, y = bt -\- ct 2 ; 
(n) x = t, y = l + t s ; 

(o) x = 1 - r 2 , y = ^ ; 

(p) x = at 2 , y = a (1 - £)' 2 ; 
(#) x = at, y = b sin t ; 
(r) x = a (1 — cos J), ?/ = a sin t ; 
(s) x = a (1 — cos £), y — b sin £. 



2. A point describes any curved path with constant speed. 
of the velocity curve ? 



What is the form 



KINEMATICS OF A POINT. CURVILINEAR MOTION 69 

3. Two points are describing free paths in one plane such that each path is the 
velocity curve of the other. If the moving points be always at corresponding 
positions, prove that the paths are conic sections. 



36. Acceleration in a plane. In plane motion velocity may be 
defined as the time-rate of change of the position vector, and the 
acceleration as the time-rate of change of the velocity vector. 

Since the velocity is a vector quantity, the acceleration is also 
a vector quantity. Let v 2 be the velocity 
vector at the instant t = t x ; and v x + Av 
the velocity vector at t = t 1 -{- At. The 
change in velocity during the interval of 
time At is represented by the vector Av and 
the average acceleration during the interval 



At is the quotient 



Av 

At' 




Velocity Curve 



The average acceleration is a vector quantity, since it is the 
quotient of a vector and a scalar. The acceleration at any instant 

(t = £[) is denned as the limit of the average acceleration — as At 

approaches zero. This corresponds to the definition of velocity 
given in Art. 35. Hence the acceleration can be obtained from 
the velocity curve in the same manner as the velocity is obtained 
from the path curve. Denoting the acceleration by f, it follows 
that its direction is the direction of the tangent to the velocity 
curve at the point corresponding to t = t r The components of 
the acceleration in the directions of the coordinate axes are given 
by formulas similar to (VIII). That is, if (x, y) are the coordinates 



of the point P on the velocity curve, then f x = — , f y 
and, since x =v afl y =v y , we have 



dy 

dt' 



(IX) 



dV. r _ d_ (dx\ _ d 2 X _ A.ti(*\ 

dt ~ dt\dt)~ dt 2 ~y K ; ' 



f = dvi, _ d (dy\ = d l y _,,, 
Jy dt dt\dt) dP T 



(*). 



The axial components of the vector acceleration are therefore 
obtained from the equations of motion by differentiating twice. 
Furthermore, a statement similar to the theorem of Art. 35 may be 
made : 



70 THEORETICAL MECHANICS 

Theorem. The axial components of the vector acceleration in 
curvilinear motion are equal to the accelerations of the axial compo- 
nents of the motion. 

The magnitude of the acceleration is obtained from its compo- 
nents by applying (I), 

/ = + vt?+77 = V(5J+ (S) 2= v ^"(os 2 + it"(OS 2 - 

37. Motion in space. The discussion of Arts. 34-36 is ex- 
tended easily to the motion of a point in space. The difference 
amounts to the consideration of the additional coordinate z. Thus 
the equations of any motion in space will have the form 

(X) » f ♦.(*), V *=+(*). *=X(«). • 

in which the independent variable represents the time. By elimi- 
nation of t from the two pairs of equations (X), the path will be 
determined in rectangular coordinates as the intersection of two 
cylinders. 

The velocity is a vector determined as in Art. 35, and if the 
axial components are v x , v y , v z , then in agreement with (VIII), 



ds 


doc 


dy 


dz 


di y 


V *=M' 


V * = M< 


v * = 7u 



(XI) 

Finally, the vector acceleration is defined as in Art. 36, and if 
fxi fy> fz are its axial components, then 

/VTT\ f - dVx — d ^ -P - (lv v - <Py * _ dVz _ d?z 

The equations of the path being given, the axial components 
of velocity and acceleration are obtained by differentiation, and 
from these components v and f are determined in magnitude and 
direction by (IV). 

38. Discussion of any motion. Given the equations of any 
motion, the determination of its characteristics involves the fol- 
lowing: 

1. Notice the nature of the component motions and draw any 
conclusions as to the general nature of the motion (periodic, etc.). 

2. Plot the path either by assuming values of t and computing 
x, y (and 2), or by eliminating t and plotting from the rectangular 
equation (or equations). Find the initial position. 



KINEMATICS OF. A POINT. CUKVILINEAE MOTION 71 



3. Differentiate and find the axial components of the velocity 
and acceleration. Determine v and/. 

4. Draw the velocity curve and discuss the variation of v with 
the time. 

5. Discuss the variation of/ with the time, both in magnitude 
and direction. 

ILLUSTRATIVE EXAMPLES 

1. Discuss the motion whose equations are 
(1) x = 2t, y=2t-±t 2 . 

Solution. Following out the discussion : 

1. The motion is not periodic. 

2. The path is a parabola. For, 

from (1), t= %x, and . •. y = x — -^ x 2 , or x 2 — 12 x -f 12 y = 0, which is a parabola. 
The initial position is the origin. 




t 


X 


y 











1 


2 


1 2 


2 


4 


2 2 


3 


6 


3 


4 


8 


Z 3 


5 


10 


If 


6 


12 





etc. 


etc. 


etc. 



3. Differentiating (1), we obtain, 
( 2 ) 



v x — — — &•> 

dt 



dy 

dt 



t. 



(3) 



.-. v = Vv x 2 + v y 2 = V8 - f t + 1 t 2 . 
s _ d 2 x _ a / _ ^ 2 2/ _ 2 



•'•f = Vf* 2 +fy 2 = 



4. The velocity curve is the straight line v x = 2. The initial velocity has the 
components (2, 2). Hence at 0, the point is moving in a direction making an angle 
of 45° with OX. The vertical 
component diminishes from 2 
when t = 0, to zero when t = 3, 
and thereafter increases numeri- 
cally but is negative. Hence 
the speed diminishes from its 
initial value vo = 2V2 to a mini- 
mum value 2 when t = 3, and 
thereafter constantly increases. 
When t = 3, the highest point 
(6, 3) is reached ; v y = 0, and hence the tangent to the path is parallel to the X-axis. 



t 


v x 


% 


v 





2 


2 


2V2 


1 


2 


H 


£vT3 


3 


2 





2 


6 


2 


- 2 


2V2 


etc. 


etc. 


etc. 


etc. 




«.-3/ 



72 



THEORETICAL MECHANICS 



5. From (3) it appears that the acceleration is constant and has a downward 
direction. 

2. Discuss the motion whose equations are 
(1) x — a cos nt, y = a sin nt. 

Solution. 1. Both axial components are periodic with the same period, namely 

Hence the moving point will return to any position in its path after an interval 

n 

of time equal to JUL and the motion is periodic. 
n 

2. Eliminating t by squaring and adding, the path is found to be the circle 

x 2 + y 2 = a 2 . 
The initial position is (a, 0). 
3. Differentiating (1), 

v x = — an sin nt, v y — an cos nt. 

. v = Vvx 2 + v y 2 = an. 

f x — — an 2 cos nt, f y =— an 2 sin nt. 

a 





Y 








^ l= 2n 


7\ 


V 







a 


t-oX 



(2) 
(3) 



Hence the speed is constant. 

f f 



4. The velocity curve is a circle of radius an, 
Also when t = the components of the velocity 
are (0, an). Hence the point describes the circle 
in a counter-clockwise direction. 

5. From (3) the magnitude of the accelera- 
tion is constant. To determine its direction, we 
observe by comparing (1) and (3) that 

(4) f x =-n 2 x, f y =-n 2 y. 

If in the figure, P is (a;, y), then the point 
(— n 2 .r, — n 2 y), lies on the line OP produced 
through O. Hence the vector acceleration at P 
is directed towards the center. 

The motion just described is called uniform circular motion. The axial com- 
ponents (1) are both simple harmonic motions with the same amplitude a and the 

— — . (Compare 




Velocity Curve 




same period 

n 

example 2, p. 49.) 



3. Discuss the motion 
whose equations are 
g (1) x = at, y=bs'mt. 

Solution. 1. The com- 
ponent of the motion in the 
direction of the I"-axis is 



periodic, while the motion in the direction of the X-axis is uniform. 



KINEMATICS OF A POINT. CURVILINEAR MOTION 73 

2. Eliminating t, the path is the sine curve 

V = b sin -, 



The speed varies between a ( when t 



etc. i and 



whose period is 2 ira and maximum ordinate is b. The initial position is (0, 0). 

3. Differentiating (1), 

(2) v x = «, v y = b cos t. .-. v = vV 2 + b 2 cos 2 t. 

(3) / x = 0, / y =-6sin*=-^. ,- f f=yff. 

4. The velocity curve is the portion of the straight line x — a between the 
points y = b and y = — 6. 

From (2) the velocity at has the components (a, &). 

IT 3 7T 

2' IP 
Va 2 + 6' 2 (when ^ = 0, 7r, etc.). That is, the speed is least 
at the highest and lowest points, and greatest at the point 
of intersection with OX. 

5. From (3) the acceleration equals the ordinate numerically but differs in 
sign. Its direction is parallel to the axis of Y. 

The motion here discussed may be thus described. The point moves with con- 
stant speed a parallel to OX and simultaneously executes simple harmonic motion 
parallel to OY. 

4. Discuss the motion represented by 
(1) x = 2 sin t, y = cos 2 t. 

Solution. 1. Both components are periodic, and it is apparent that the 
moving point will return to any position in its path after an interval of time equal 
to 2tt. 




t 


X 


y 








i 


\* 


2 


-i 


TT 





i 


I* 


-2 


-i 


2tt 


o 


i 


etc. 


etc. 


etc. 




2. Since cos 2 t = 1 — 2 sin 2 t, we find, on eliminating t, 

y = 1 — 2 sin 2 t = 1 — \ x 2 . 
That is, the path is a portion of the parabola x 2 + 2 y — 2 = 0. The initial position 
is 4(0, 1). 

3. Differentiating (1), 

(2) v x — 2 cos t, v y = — 2 sin 2 t. 



(3) 



.*. v = 2 Vcos 2 £ + sin 2 2 £. 

/a = — 2 sin J = — x, f v = — 4 cos 2 « = — 4 y. 

.-. /= Vac 2 + K>3y" 2 . 



74 



THEOKETICAL MECHANICS 



4. The velocity curve, plotted from the parametric equations, has the form of 
the figure 8. 

From (2), when t = 0, v x = 2, v y = 0. The point ini- 
tially at A moves to the right to the extreme position 
B (2, — 1), at which point (t =| ir) the velocity is zero. 
It then returns through A to C {— 2, —1), at which 
point v is again zero (t = f ir). The point is again at A 
when t = 2 7r, and the vibration is then repeated. 




t 


v* 


Vj, 


V 





2 





2 


i* 











IT 


-2 





2 


I 71 " 











2tt 


2 





2 


etc. 


etc. 


etc. 


etc. 



t 


/. 


/» 


/ 





. 


-4 


4 


1 7T 


-2 


4 


v'20 


7T 





-4 


4 


«» 


2 


4 


V20 


2tt 





-4 


4 


etc. 


etc. 


etc. 


etc. 



5. From (3), the acceleration has the components (— x, —4?/). At A the 
acceleration is downwards ; at B or C it is tangent to the path. For, differentiat- 
ing the equation of the path in 2, we get — ^ = — x, and hence the slope at B is 

dx . 
—2. But the slope of the vector whose components are (—2, 4) is =—2, 

— A 

Therefore the acceleration at B is tangential. Similarly for C. In the figure 
the vector acceleration is drawn to scale. 

AA.', BB', and CC are the vectors representing the acceleration at the points 
A, B, C, respectively. The unit of length for the acceleration vector is f the unit 
on the X-axis. 

The motion just discussed is therefore an os- 
cillation with parabolic path, the period being 2 ir. 
The components (1) are simple harmonic motions 
with different periods, namely, 2 ir and ir. Their 
resultant motion is that of a point executing simul- 
taneously simple harmonic motions parallel to 
perpendicular axes, the ratio of the periods being 2. 

5. Discuss the motion in space defined by 

(1) x = a cos t, y = a sin t, z = bt. 

Solution. 1. The x- and ^/-components of the 
motion are harmonic vibrations, and the 0-compo- 
nent is uniform motion. 

2. The path is a helix on the cylinder x* + y 2 = a 2 (Calculus, p. 272). The 
initial position is (a, 0, 0). 




KINEMATICS OF A POINT. CURVILINEAR MOTION 75 

3. From (1), we obtain, by differentiation, 
(2) v x =—asint, v y =acost, v z =b. 



. -. v = Va 2 + V\ 
(3) f x = — a cos t = — x, f y = — a sin t = — y, /., = 0. 

• ••/=«. 

4. When t = 0, v x = 0,v y = a, tfe = &. 

Hence (6 > 0) the point describes the helix with constant speed in the upward 
direction. 

5. The acceleration is constant in magnitude, is parallel to the XT-plane, and 
is directed towards the Z-axis, since the direction from (0, 0, 0) to (— x, — y, 0), 
when drawn from (x, y, z), will pass through the axis OZ. 

By comparison with example 2, it is seen that (1) may be regarded as the 
motion of a point having simultaneously uniform circular motion around OZ and 
constant speed along OZ. Such a motion is obviously that of any point on the 
periphery of a screw which is forced inward at constant speed. For this reason 
the motion defined by (1) is called a screw motion. 

PROBLEMS 

1. .Discuss each of the following motions : „ 

(a) x = St, y = 2 — t; (I) x = a (t — sin t), y = a (1 — cost); 

(b) x = 1 — St, y = 6 + t ; (m) x = a cos 3 1, y = a sin 3 1 ; 

(c) x = a + bt, y — c + dt ; (n) x = a {t + sin£), y = a (1 — cost); 

(d) x = t 2 , y=\t\ io) x = a(t — sint), y=b(l — cost); 

(e) x = 1 — t, y ;= t 2 ; (p) x = a (t + sin t) , y = b (1 — cos f) ; 
(/) x = St, y = Qt ■<- £ 2 ; (g) cc = asin 3 £, ?/ = &cos 3 £; 

(gO x=at, y = bt—%gt 2 ; (r) x = at 2 , y = a (1 - t) 2 ; 

(h) x = at 2 + bt, y = ct ; (s) x = a (1 — cost), y = asmt; 

(i) x = t, y = t z ; (t) x = a (I — cost), y = bsint; 

(j) x = t 2 , y = t 3 ; (u) x = cos t, y = 4 sin \ t ; 

(k) x = ae kt , y — be~ kt ; (v) x — a cos t, y = a cos 2 1 ; 

(w) x = a sin 2 t, y — a sin t. 

2. Discuss each of the following motions, the components in each case being 
simple harmonic motions : 

(a) x = 2 sin t, y = 2 cos t; (e) x — a sin t, y = b sin (£ + /3) ; 

(6) sc = 2 sin £, y = 3 cos t ; (/) as = 2 sin a £, y = cos t ; 

(c) x = sint, y =cos2t; (g) x =acost, y = bcos2t; 

(d) x = a sin fot, y = b cos At ; (h) x = sin 1 1, y = a sin t ; 

(i) x = acos (Jet + /3), y = & sin (At + /3) ; 
(j) z = acos (AZ + /3), y = & cos(At + /3). 

3. Discuss each of the following motions in space : 

(a) x = t, y = t + l, z = 3-t; 

(b) x = l-2 t, y = 2t- 5, z=t-6; 

(c) x — at, y =bt, z = ct ; 



76 THEORETICAL MECHANICS 

(d) x = at + «i, ?/ = bt + &i, « = ct + Ci ; 

(e) x = sin £, «/ = £, z = cos £ ; 

(/) x = bt, y = a sin t, z — a cos t ; 

(<7) x = a cos £, y = bt, z = asint ; 

(h) x = a cos t, y = 6 sin t, z = A cos t + B sin £ ; 

(0 x = t, y = 1 — t 2 , z = 3 1 2 + 4 « ; 

( j) x = £ 2 + 8 1 + 1, y = t 2 -Z, z — 1 - '8 t; 

(k) x = 2 cos t, y = 3 cos £, = i ; 

(Z) x = sin t, y = cos 2 £, = sin t ; 

(m) x = sin t, y = — - cos t, z = — - cos£ ; 

V2 V2 

(n) x = acos(kt + /S), y = ftsin (to + £), 2 = £; 

(0) x = a cos 3 £, y = bt, z = a sin 3 £ ; 

(p) x = a (t — sin £), 2/ = t, z ~ a (1 — cos £)-. 

39. Motion in a prescribed path. The question may be raised: 
What characteristics must any motion on an ellipse possess ? 
Certain points are readily settled. If the path is 

(1) b 2 x 2 + a 2 y 2 = a 2 b\ 

either axial component of the motion (VII) may be chosen, and 
the other is then determined. Thus, if we choose the x component 
as the simple harmonic motion, 

x = a cos Jet, 
then, from (1), by substitution, 

a 2 b 2 cos 2 kt + a 2 y 2 = a 2 b 2 , or 
y = b sin kt. 

In general, on a prescribed path one axial component may be 
chosen arbitrarily, and the other is then found by substitution and 
solving. That is, we set x = <f>(t), where <£(£) is assumed, substi- 
tute in the given rectangular equation, and solve for y. 

Further useful equations are the following : 

From v x = — , v v = -#-, we obtain 

dt dt 

dy 

(2) dy = dt == v Jl ^ 
dx dx v x 

dt 



KINEMATICS OF A POINT. CURVILINEAK MOTION 77 
When the path is given, -^ is found by differentiation, and (2) 

(XX 

gives a relation between the components of the velocity which 
holds for each point of the path. For example, for the ellipse (1), 
this relation is 

<&y v x ' 

Differentiating (2) with respect to z, we get 

dv y _ dv x 
<Py = d_fv^\ ^_dx = Vx dt ~ Vy ~dt ^ 

dx 2 dt\vj ' dt v/ 

In this equation the value of the second derivative of y with re- 
spect to x is found from the equation of the prescribed path. 

From (2), it has been seen that one of the axial components of 
the velocity may be chosen and the other is theft determined. 
Knowing v x and v y , we may obtain f x and f y by differentiation, 
and then check the results by equation (3). 

Illustrative Example. If the path is the equilateral hyperbola xy = c, and 

v y = k (constant) , find v x and f x . 

Solution. From the equation of the path, -^ = '— ¥ • 

dx x 

Hence, from (2), 

(4) *=*+*= -fe 

dx y 

From the equation of the path, we find y = -■ 

x 

By substituting, (4) becomes 

(5) v x = - -x\ 

c 

Differentiating with respect to t, 

(6) /. = - 2 -^xv x = 2 -^x*. 

c c 2 

c (Pu 2c 

From the equation of the path y = -. we find — £ = — , and substituting in (3) 

x dx 2 x s 

from (5) and remembering that/,, = 0, the results check. 



78 THEORETICAL MECHANICS 



PROBLEMS 

1. In the following problems the path is given. Find (1) f x if v y = (3 ; (2) f y if 
v x = a. 

(a) xy=a 2 . Ans. ^x 3 ; —y*. 

a 4 a* 

(b) y — a x . , Ans. "~ ^ — ; a 2 (log a) 2 w. 

a 2 * log a 

(c) y 2 = 4 ax. ^Iws. ' 



2a' 



a 2 b 2 



w^.-S=i. 

a- & 2 



(Z)^ 2 ' +2/ 2l = a"2; 



■y i / c?x w \ 
(a) x = a arc vers - — (2 ay — ?/ 2 ) 2 . ( Here — = r ) 

a V dy (Zay-y*)* 1 



Ans. 



ace? 



(*)y = f(^+.« •)■ 



.Ans. — 



(2ay-y*y y 

■ay . a 2 ?/ 



2. In the following problems the path and the component of velocity along 
one axis are given; to find the component of acceleration along the other axis. 

(a) as+-y = l; e s = cos &£. Aws. /y^&sin^. 

(6) Ax + 5*/+C = 0; v x = t 2 -t. Ans. f v =—(l-2f). 

(c) ?/ 2 = 4ax; v y = ct. . Ans. f x = -^-(ct 2 + y). 

(d) y 2 =4ax; v x =smt. Ans. f y = -^(?/ 2 cos t — 4 a 2 sin 2 t). 



a 



a \ 2 y cos 2 * , a 3 sin 2 2 « 
I 



(e) x 2 + */ 2 = a 2 ; t,, = |dn2t. ^n«. /. = -||.J— + 

2 £x <x 2 £* 
(/) x 2 + 2/ 2 = a 2 ; *. = *. 4w*. / r = -jg" 

x 2 ?/ 2 
(a) -g + '— = 1 ; v y = nb cos n£. 

K ' a 2 b 2 b 2 \x x 3 / 

0') ^ - ?-, + 1 = ° ; v * = a sec2 '• 

v a' 2 6 2 

(A:) xy = a 2 ; v x = asec 2 t. 



KINEMATICS OF A POINT. CURVILINEAR MOTION 79 

3. In the following problems the path and one component of the motion are 
given ; to find the components of velocity and acceleration and to discuss the 
motion. 

(a) y 2 = 4ax; y = ct. {e) ^_t =:1; x = ct 

a 2 b 2 



(/) ±_ _ SL + l = ; x = a tan t 



(<0 


X 2 

a 2 ' 


y2_ 

b 2 


:1. 


(/) 


xy 


= a 2 . 




(9) 


i 

X* 


+ y*-- 


1 

= a 2 



(b) y 2 = 4 ax ; x = a cos t. 

(c) x 2 + y 2 = a 2 ; x = b cos w£ ; (b fL a). 
(<Z) z 2 + ?/ 2 = a 2 ; y - ct 2 . " ' a 2 b 2 

(g) xy = a 2 ; y = a tan t. 

4. A point describes the curve given with constant speed ; to determine the 
components of velocity and acceleration. 

(a) Ax + By+ C=0. 

(6) x 2 + y 2 = a 2 . 

(c) y 2 = 4 ax. 

2 2 (h)y = a log sin x. 

a 2 b 2 (i) y = & log cos x. 

5. Given v x = kx, v y =ky. Show that the path is a straight line passing 
through the origin, and find the components of acceleration. 

6. Given v x = ky, v y = kx. Find f x , / y , and the equation of the path. 

7. A wheel rolls on a horizontal plane so that its center has constant speed. 
Compare the speed at any instant of a point on the circumference with the speed 
of the center. 

8. Find the axial components of the acceleration in problem 7, show that the 
acceleration at any instant of a point on the circumference is constant in magnitude 

and is directed towards the center of the wheel. 



(=-:) 



9. A wheel rolls upon the inside of a second wheel whose diameter is twice its 
diameter. If the center of the smaller wheel moves with constant speed, show that 
a point upon its circumference will execute simple harmonic motion. 

10. The pin of a crank moves in a groove in a vertical bar whose extremities 
move in horizontal grooves. If the crank pin rotates with constant speed, show 
that any point of the vertical bar will execute simple harmonic motion. 

11. A point describes a curve with an acceleration parallel to OY. Show that 

f — c 2 — \ , where c is the constant speed parallel to OX. 
ax 2 

12. A particle describes the cycloid x = a{6 — sin0), y = a(\ — cos0). Show 

2 v 2 
that v 2 = — Z- . If the acceleration is at right angles to the line joining the cusps, 

y 

show that it varies inversely as the square of the distance from this line, or also 
directly as v 4 . 



80 



THEORETICAL MECHANICS 



13. A point moves on the catenary y = % a (e a + e a ). 



Show that 



v x *y*. 



If the acceleration is parallel to OF, show that it varies directly as the velocity. 

14. What points on the rim of the wheel in problem 7 have the same speed as 
the center ? Ans. Points of an arc of 60° described about the lowest point. 

40. Tangential and normal accelerations. For plane motion 
the components of the acceleration vector in the directions of the 
coordinate axes are given by (IX). The components in the direc- 
tions of the tangent and normal to the path, are obtained by- 
applying (II). 

We first adopt a convention as to the positive direction along 
the tangent and normal. The positive direction along the tangent 
PT shall agree with the direction of the velocity. The positive 
direction along the normal PN shall agree with the direction 
obtained by rotating PT counter-clockwise through a right angle. 
Hence, by the definition, 

" (*, iVO =!+(*, T). 

If f t and /„ are the tangential and 
normal components of f, from (II), 

(2) /, =f x cos (x, T) +f y sin (x, T). 

(3) /„ =/, oos O, JV) +/, sin O, JV). 
The second member of (2) is re- 

Since by assumption (x, T) = (a?, v), we have 

cos (a;, T) = cos (a;, v~) = — ; 




duced as follows. 



eo 



sin (#, T) = sin (#, v) = - M \ 



Hence, by substitution in (2), using (IX), we get 



/, 



dv x 
~dt 



v r , dv,. 

_x _| y 

v dt 



= - v 



dv 



dv. 



dt + y dt 



dv 
dt 



Since v 2 



+ Vy 2 , by differentiation, 



2v^ = 2v x ^ + 2v/^ 
dt dt dt 



^ = - v. 



dv x 
dt 



)■ 



KINEMATICS OF A POINT. CURVILINEAR MOTION 81 
Similarly, to transform (3), we have, from (1) and (4), 

cos (a, JV) = - sin (x, T) = - ^/, 

v 

sin O, JV) = cos (a;, T) = ^. 
Substituting in (3), we obtain 
(5) f = v *fy~ v vf* . 

If R denotes the radius of curvature of the path (formula 
64, Chapter XIV), we have, by (VIII) and (IX), 











R 


v 3 






v ,fy ~ v 


yJx 


Hence, 


we 


write 


(5) 


in 


the form 




(6) 










V 2 

f =—■ 





For reference later we give also another form iov f n . From 
(3), Art. 39, we have 

&y = v xf„- v ufz . 
dx 2 v/ 

Hence, from (5), 

(7) ' /„ = ^f|- 

v dx A 

The results found give the 

Theorem. If the vector acceleration at any point of the path is 
resolved along tangent and normal, its components are 

where R is the radius of curvature. 

Since f t and/ w are at right angles, we have obviously 



f-Yft+ff- 

Two important results follow from (XIII). 1. If the path is a 

straight line, — = 0, and .'.f n = 0. That is, in rectilinear motion 
R 

the vector acceleration is directed along the path. 2. If the speed 

ds d 2 s 

is constant, — = c, whence — -= and .-. f t = 0. Hence in curvi- 



82 



THEORETICAL MECHANICS 



fn 



linear motion with constant speed the acceleration is directed along 
the normal. 

Furthermore, in any curvilinear motion (/„^=0), the accelera- 
tion is directed towards the concave side of the path. To show this, 
four cases must be considered. From formula (7), for f n it is 

plain that /„ is positive or negative according as v x and — ^ have 

ax 

like or unlike signs. By Calculus, p. 137, the path is concave up- 
wards or downwards according as — &- is positive or negative. The 

doer 

four cases to be considered are : 

1. The path is concave, upwards and the point is moving 

towards the right. Therefore, — ^ and v x are positive ; hence f n is 

dor 

positive and the resultant of f t and /„, that is, f (fig. a) is di- 
rected towards the con- 
cave side of the curve. 

2. The path is con- 
cave upwards and the 
point is moving towards 
the left. Therefore, 

dx 2 

is negative ; hence f n is 
negative. By definition 
(1), the normal PN is 
directed downwards, hence /„ is directed upwards and the re- 
sultant f is directed towards the concave side of the curve 

(%. j). 

Similar results follow for the two cases when the curve is 
concave downwards, as in figures c and d. 

Since the direction of the tangent agrees with the direction of 
the velocity, we have from the figures the criterion : The velocity 
vector is rotating counter-clockwise when f n is positive, clockwise when 
f n is negative. 

The significance of the algebraic sign of f t is easily determined. 

d fds\ 
Since f t = — (— -), it is seen that when f t is positive the speed is 
dt\dtj 

increasing; when negative, the speed is decreasing. 



r ft 
(a) 



N 




, / 



(C) 




is positive and v a 



KINEMATICS OF A POINT. CURVILINEAR MOTION 83 



When the equations of motion are given, we proceed as fol- 
lows to find f t and f n : 

1. Differentiate and find v x , v y , f x , f y . 



2. Find v from v = Vv x 2 + v y 2 . 

3. Differentiate this last result, giving f t = 

4. Find f n by (5), p. 81. 



dv 
~di 



Illustrative Example. Determine the normal and tangential accelerations 
in the motion denned by 

x = a cos t, y = b sin t. 

Solution. Eliminating t, the path is the ellipse 
b 2 x 2 + a 2 y 2 = a 2 b 2 . 

Following the directions given, we find 
v x = — a sin t , v y = b cos t, f x = —a cos t, f y = —b sin t. 




Hence v = + Va 2 sin 2 £ + b' 2 cos 2 £. From these values we obtain 
(fo (a 2 — b 2 ) sin £ • cos t 



ft = 

fn 



dt + va 2 sin 2 £ + b' 2 cos 2 £ 
^y/v — v„f x a& 



z 


X 


2/ 


/• 


fn 





a 








a 


*» 





6 





b 


7T 


— a 








a 


i* 





-b 





b 


2tt 


a 








a 



) + Va 2 sin 2 t + b 2 cos 2 £ 

We note the following table of values. The point 
describes the ellipse counter-clockwise. The normal 
acceleration is always positive, agreeing with the fact 
that the velocity vector rotates always counter-clock- 
wise. Since f t > when the point lies in the first and 
third quadrants, the speed increases from A to B and 
A' to B'. Similarly, from B to A' and B' to A the 
speed decreases. 



PROBLEMS 

Find/, and/„ for problems 1 and 2, p. 75, and discuss the results. 

41. Equations in polar coordinates. In many cases it is more 
advantageous to employ polar coordinates in studying motion in 
a plane. If (jo, 0) are the polar coordinates of a moving point _P, 
the equations of motion have the form 

(xiv) p =+(*), e = +(*), 

since obviously p and 6 are now functions of t. 

In rectangular coordinates the derivatives of x and y with 
respect to the time were of fundamental importance. Similarly 
in using polar coordinates we shall expect their derivatives to 



84 



THEORETICAL MECHANICS 



appear. The time-rate of change of the radius vector p is called 
the radial velocity of the point (/a, #). The time-rate of change 




of the vectorial angle 6 is called the 
gular velocity co of the point (/o, 
That is, 

dp 

~dt 

— — (o = angular velocity. 



an- 
6). 



(i) 



radial velocity, 



We desire to obtain the components 
of the velocity vector and of the acceleration vector when re- 
solved along and perpendicular to the radius vector. We first adopt 
a convention as to the positive directions along these lines. 

The positive direction along the radius vector is defined as in 
Analytic Geometry, p. 149. The positive direction perpendicu- 
lar to the radius vector is the direction obtained by increasing the 
vectorial angle 6 by a right angle. 

Denoting the components of the velocity and acceleration 
vectors along the radius vector by v p and f p and perpendicular to 
the radius vector by vq and f& respectively, we have, (apply- 
ing(II)), 



v p = v x cos 6 + Vy sin #, 



vq = v x cos ( ^ + 6 J + v y sin ( ^ -f- 6 ] = — v x sin 6 -f v y cos 



(2) 

and 

f/p = fx c °s + f y sin 6, 

(8) j/'=/* C08 (f + °) +/* 8in (f + °) = ~ f** in0 + /, cos ^- 

To transform the derivatives of the rectangular coordinates into 
the derivatives of the polar coordinates, we have the relation, 

\ y = p sin 6. 
By differentiating (4), we obtain 



(5) 



/) dp . ad6 

v r = cos v -£ — p sin 6 — , 
dt r dt 

■ ndp , n d0 

v.. = sin 6 -t- -f p cos 6 — , 
y dt r dt 



KINEMATICS OF A POINT. CURVILINEAR MOTION $5 
and 



(6) 



f r — cos — £ — 2 sin -^ • - p cos 01 — — p sm — , 

Jx dt 2 dt dt v \dtj * dt 2 

* ■ a d2 P , o adp d6 . a fd0\ 2 , a d?0 



Substituting (5) and (6) in (2) and (3), respectively, we 
obtain, after simplifying, 



(XV) 



<f 2 6 , dp dQ ld. M 

/. r=0 + 2— *- • = (P<0) 

Jd v dt 2 dt dt pdt yv J 



since 



1 d 



^0*») = 



pdt 



ld_f 2 d6) = dM.Qdp dO 
pdt\ P dt) P dt 2 ~ dt dt 



Of course v = Vv p 2 + v e 2 , f= V/ p 2 + f e 2 , as usual. 



Illustrative Example. A point describes a circle whose equation is given 
in polar coordinates. Discuss formulas (XV) for this case (compare Art. 38). 

Solution. If the origin is on the circumference, the equation is 

(1) 4 P = 2a cos 6. 

Differentiating with respect to t, 



(2) ^ = _2a sin 
dt 



06 

dt 



or v n = — 2 a sin 



v 2 = v 2 



v B 2 = 4 a 2 sin 2 6 ■ ufl + p 2 w 2 
= (4 a 2 sin 2 6 + 4 a 2 cos 2 0> 2 . 
Hence 
(3) v 2 = 4 a 2 w 2 , or u 



2a 



and v p =—v sin 




Equations (3) express angular and radial velocity in terms of speed, and are 
easily found directly from the figure. This verification is left to the student. 
To find the component accelerations, differentiate (2) again. This gives 



ivp = -2aeosew 2 -2asm° du} 



dt 



dt 



2 a sin 6 



dt 



86 THEORETICAL MECHANICS 

(4) .-./ p=-2p«a-2asintf^? 

dt 

o o , v p du 
w dt 

- _ 2 pw 2 + ^ a (by (4), Art. 42). 

CO 

Similarly, 

(5) / 9 =--(^) = 2^ + p- 

= 2 wv p + poc. 
Substituting in / 2 =/ p " 2 +f$ 2 , we find after reducing, 

(6) /2=W4 w 2 + «?\ = 4 a 2(4 w 4 +a 2). 

This equation expresses the total acceleration in terms of the angular velocity 
and acceleration. 

In particular, assume fg = 0, that is, let the acceleration be directed towards 
the origin. Then, from (5), 

(7) A (p 2 w) = o. ... p2 w = C) and w= c 

dY p 2 

Also, from (5), 

2 wv_ 
2 uv p + pa = 0. .. a = — — e • 
p 
Then (6) becomes 

/2 = d g8 / 4 ^ , 4 "V\ = 16 aVjg = 64 aM 

\ P 2 / P 2 P 2 



(8) .-./=- 



8 a 2 a> 2 8 a 2 c 2 



P P" 

the negative sign being used since the acceleration must be directed towards 0. 

This result is due to Newton, and may be stated as follows : If a particle 
describes a circle with an acceleration directed towards a point on the circumference, 
the acceleration must be inversely proportional to the fifth power of the distance. 

PROBLEMS 

1. Plot the path,* find v P , vq, f p , fg, and discuss the motion defined by the 
equation : 

(a) p = 2 a sin t 2 , 6 = t 2 ; 

(6) p = 2at, = arc cos £ (Q^t'< 1) ; 

(c) p = a cos t, 6 = a sec t ; 

(d) p = a sin t, B — sin t ; 

(e) p = tan t, 6 = cot t ; 
(/) p-a tan t, = cot 2 1 ; 
(?) p = e*«, = £; 

(h) p = a'(l — 0» = arccosZ (0^£^1) ; 
(i) p = a sin £, 6 = It; 

(j) p = a cos £, # = i £• 

* The path may be plotted from the parametric form as given, or the ordinary polar 
equation may be obtained by eliminating t. 



KINEMATICS OF A POINT. CURVILINEAR MOTION 87 



ep 



2. A point describes the ellipse p - 

1 — e cos t 

Let 6 be given in terms of E by the relations, 



sinfl = Vl - e2 - sin ^ cos0= ^sE+e 



1 + e cos E 



1 + ecosE 



and E be given in terms of t by 



nt = E + e sin E. 



Prove 



f P = - *ip /• = 0, where a = -S£_. 
p 2 1 — e 2 



42. Rotation. When the path is a circle, the motion is called 
rotation. If the radius is r, the equations of motion are 

(1) p= r , 6 = f(t-). 

The position of the point is completely de- 
termined if 6 is known. For this reason, 
the equation p = r is unimportant and it 
is customary to call the second equation, 

(2) *-*(*), 

the equation of the rotation. 

From (2), we obtain by differentiation, 

cW 

dt 




(3) 
(4) 



angular velocity = co ; 



d 2 day 7 7 

-— = — - -= angular acceleration = a. 

dt 2 dt J 



That is, the angular acceleration is the time-rate of change of 
angular velocity. Angles being measured in radians, angular 
velocity is measured in radians per second. For example, if 

a? = 1 7T and is constant, the radius OP 
rotates through a right angle in each sec- 
ond. In the same way angular accelera- 
tion is measured in radians per second in 
each second. For example, if a = 1 and 
is constant, the radius OP rotates with 
increasing angular velocity, the gain be- 
ing one radian per second in each second. 
The speed in rotation is readily found. 
For if 6 — is the angle turned through in the time £, and s the 
length of the corresponding arc, we have 

s = r(6- O ). 




88 THEORETICAL MECHANICS 

Hence, by differentiation, 

^r N ds dO 

(o) — = r— r, or v = rco. 

v J dt dt 

In rotation the speed equals the angular velocity times the radius. 

Next, consider the tangential and normal accelerations. These 
are also readily expressible in terms of co and a. For by (XIII) 
and (5), 



(6) 



/. dv dco 
f t = — = r--= ra, 
dt dt 

v 2 y.2^2 

fn- ~s = =ra> 2 . 

Jt r 



The same are found from (XV) by noting that in rotation 

IP == ~Jni Je ~Jt' 

Theorem. If co and a are respectively the angular velocity and 
angular acceleration in rotation, the speed and acceleration are 
determined from 

(XVI) v = rco, f t — ra, f n = rco 2 . 

Illustrative Example. A fly wheel "is making 120 revolutions per minute 
(R.P.M.). If the angular velocity diminishes at a constant rate, find the number 
of revolutions if the wheel stops in one minute. 

Solution. The motion of the wheel is determined by the motion of one of its 
points. Let w be the initial angular velocity. 

Then, since * 120 R.P.M. = 2 R.P.S. = 4 w radians per second, we have w = 4 w. 

Since the angular acceleration is constant, 

a = — = Jc, .-. w = kt + c, where c is the constant of integration. But w = wo 
dt 

when t = 0. 

(1) . •. (o = kt'+ w or a = M + 4 7T. 
Since the wheel comes to rest in 60 sec, o = when t = 60. 

. \ = 60 k + 4 7T, and k = - ^ w. 

(2) .-. co = -fg- 7r£ + 4 7r. 

dd 
Writing w = — , integrating and assuming 6 — if t — 0, we obtain from {2) 
dt 

(3) 6 = - 3V TTt 2 + 4 TTt, 

which gives the angle turned through in any time. If t = 60, 6 = 120 tt, and hence 
the number of revolutions is — = 60. 



* In general, angular velocity = — • R.P.M. 



KINEMATICS OF A POINT. CURVILINEAR MOTION 89 



PROBLEMS 

1. In the following problems the equation of motion of a point describing a 
circle is given. Discuss the motion. 



(a) 


= at + & ; 


(6) 


= at 2 + &£ + c 


(<0 


= sin £ ; 


(d) 


e = t 3 - t\ 


00 


2 



2. A fly wheel making 360 R.P.M. is subject to a constant retardation of 1 
radian per second per second. How many revolutions does it make before stop- 
ping ? What time is required ? Ans. 36 w revolutions ; 12 ir sec. 

3. A fly wheel starting from rest is subject to a constant angular acceleration 

of \ radian per second per second for two minutes. Find the angular velocity and 

the number of revolutions made at the end of the first minute; at the end of the 

second minute. . 900 ^ -d ^ 450 1800 „ ^ ,., 1800 _„ 

Ans. R.P.M., — rev.; R.P.M. , rev. 

7T 7T 7T 7T 

4. A fly wheel starting from rest and subject to a constant angular acceleration 
for 3 minutes makes 5000 revolutions. Find the acceleration. 

Ans. a = f — - rad. per second 2 . 
81 F 

5. A fly wheel making 500 R.P.M. and subject to a constant retardation comes 
to rest after making 2000 revolutions. What time is required ? Ans. 8 min. 



CHAPTER IV 

KINETICS OF A MATERIAL PARTICLE 

43. Momentum. In the preceding chapters motion of a material 
particle has been studied without reference to mass or force. The 
latter are now to be taken into consideration. We begin with the 
definition : 

Momentum or quantity of motion is the product of mass and 
velocity, or 

(1) Momentum at any instant = mv; 

From the definition it is plain that momentum is a vector quan- 
tity, being the product of the vector velocity by the positive num- 
ber m. The direction of the vector momentum is the same as that 
of v, but its magnitude equals the product of mass and speed. 

44. Force. The science of Mechanics is founded upon laws 
or axioms which sum up the results of experience in the observa- 
tion of motion. A set of three Laws of Motion was proposed by 
Sir Isaac Newton (1642-1727), the statement of which is general 
enough for present purposes. Considering these laws as needed 
in the development of our subject, we begin with the 

First Law op Motion. Every body persists in its state of 
rest or of uniform motion in a straight line, except in so far as it 
may be compelled by force to change that state. 

Remembering that uniform motion in a straight line means 
motion with constant vector velocity, it is plain that uniform motion 
means constant vector momentum. The First Law is often ex- 
pressed by saying that the body has inertia. A body has no power 
of itself to change its state of rest or motion, but continues to 
move with constant momentum when not acted upon by an im- 
pressed force. That is, by the First Law we conclude that no 
force is acting upon a body if the body is at rest or moving with con- 
stant momentum. 

If, however, the momentum is variable, then the existence of 
forces acting upon the body is inferred. We thus come to the 

90 



KINETICS OF A MATERIAL PARTICLE 91 

Second Law of Motion. Change in momentum is caused 
by forces acting upon the body. Force and change in momen- 
tum agree in direction, and the magnitude of the force at any 
instant is proportional to the time-rate of change in momentum. 

In this statement of the Second Law is contained the defini- 
tion of force. For consider the motion of a material particle 
of mass m. Its momentum at any instant equals mv. Since m 
is constant, change in momentum means change in vector velocity, 
and the direction of change in velocity we know agrees with the 
direction of the acceleration. By the Second Law, therefore, 
force and acceleration agree in direction. Furthermore, the 
magnitude of the force at any instant is proportional to the time- 
rate of change in momentum; that is, 

(2) Force at any instant = k — (mv)= km — = kmf,* 

dt\ J dt 

where k is a constant factor of proportionality. Hence the Second 
Law leads to the result : 

Tlie force acting at any instant upon a material particle has the 
direction of the vector acceleration and in magnitude is proportional 
to the product of the mass and acceleration. Force is therefore the 
cause of acceleration. 

The value of the factor k in (2) depends upon the units 
assumed. Evidently for analytical purposes it is convenient to 
assume k = 1. This is shown below to be equivalent to assuming 
that force is measured in so-called scientific units. For theoretical 
purposes, therefore, we may assume as the magnitude of force, 

(I) Force = m ^ = mf. 

dt 

In Applied Mechanics, however, it is found more convenient to 
select k not equal to unity. (See Art. 45.) 

Observation of falling bodies makes familiar the phenomenon 
of changing momentum. The force in question is then called 
the weight of the body, or also, the force of gravity. That is, 
weight is the force of attraction exerted by the earth upon other 
bodies. The acceleration caused by weight is nearly constant in 

* In equation (2) the differentiation is made on the assumption that the mass is 
constant. If the mass is variable, a special investigation is required. See Routh, 
Dynamics of a Particle, p. 80. 



92 THEOEETICAL MECHANICS 

a small region near the earth's surface and is denoted by g. This 
acceleration is also called the intensity of gravity. The numerical 
value of g varies from place to place and also depends upon the 
units of length and time adopted. In the English and French 
systems, respectively, as an average value, 

g = 32.2 ft. per sec. in 1 sec. (English), 
g= 983 cm. per sec. in 1 sec. (French). 

Dimensions. From the definition of force it follows that its 
dimensions are mass times acceleration. The derived unit of 
force is therefore expressed in terms of the fundamental units of 
mass, distance, and time by the dimensional equation 

Force = mass x len g th 
time 2 

45. Units of force. Scientific units. For theoretical purposes 
it is convenient to define unit force as that force which will 
produce unit acceleration in unit mass. With this definition it is 
apparent that in equation (2), Art. 44, the factor of proportion- 
ality, &, is unity. Hence, in scientific units, 

(1) Force = mass times acceleration. 

In the English system, the unit of mass is the pound and 
the scientific unit of force is the poundal. Hence, one poundal 
is that force which will give to a mass of one pound an acceleration 
of one foot per second in one second. In the French system, the 
unit of mass is the gram and the unit of force is the dyne. 
Hence, one dyne is that force which will give to a mass of one 
gram an acceleration of one centimeter per second in one second. 

Technical units. In engineering practice the English unit of 
force is equal to the weight of unit mass and is called the pound. 
Referring to (2), Art. 44, since the force in question is weight, 
we must replace /by g, and thus obtain 

F = kmg. 
By hypothesis, when m is unity, so also is F, 
.-. 1 = kg and . •. k = 1 4- g. 

Substituting in (2), Art. 44, gives as the magnitude of force in 
technical units, 

(2) Force = mass times acceleration divided hy g. 



KINETICS OF A MATERIAL PARTICLE 93 

Comparison of the two systems of units. Mass, time, and 
length are measured by the same units in both systems. As just 
explained, however, force is measured by different units. To 
find the relation between the latter, we may apply (1) to the 
case of weight, whence, in scientific English units, 

(3) Weight = rag (poundals). 

Since, by definition of the technical unit, the weight of a 1-lb. 
mass equals 1 lb. of force, hence the equivalence, 

(4) One pound of force = g poundals. 

The student will observe that in technical units weight and 
mass are numerically equal. The difference is one of dimensions 
only. 

The following Table of Equivalents, together with equation 
(4), will be found useful : 

English French 

1 foot = 80.48 centimeters; 

1 pound (mass) = 453.6 grams ; 
1 poundal = 13,825 dynes; 

1 pound (force) = 4.45 (10) 5 dynes. 

46. Rectilinear motion. If the path of a material particle is 
a straight line, the expressions for the acceleration are given by 
(III), Chapter II. Hence, applying (I), and denoting the force* 
by F, we have 

-,_. oV'x dv dv 

Dividing by w, we have the force equation or the differential 
equation of motion in a straight line. 

(II) («) ^ = §f. or W *M , or (c) ^=v%», 
v y m dt 2 m dt m doc 

Suppose the mass m is given and the force is known. It 
is required to discuss the motion. For this purpose we must 
determine x from equations (II) by integration. If F is a function 
of the time only, (a) should be used ; if F is a function of the 

* The discussion of the text assumes scientific units in all cases. 



94 THEOEETICAL MECHANICS 

velocity only, (5) should be used ; and if F is a function of the 
distance only, it is usually more convenient to use (<?). However, 
in case of a linear function of v and x, that is, if F = Av + Bx + (7, 
where A and B are constants and C is constant or involves £, use 
O) (see equations 71, 72, 73, 74, Chapter XIV). If F is a con- 
stant, either form may be used. 

The force alone is not sufficient to determine the motion 
completely. For example, let us consider the case of a particle 
projected vertically in a vacuum. Obviously the motion will 
depend upon the position (on the vertical line OX} from which 
the particle is started, and upon the velocity with which it is 
projected. The initial position x and the initial velocity v are 
called the initial conditions, and it will be shown that when known 
they determine the motion completely. The only force acting 
is the weight, whose magnitude is mg. The direction of the force 
is downward, and if we choose the positive direction along OX 
downward, we have, from (II), (6), 



or 



Multiplying by dt and integrating, 

(1) v=gt + c v 

* . dx 

where c x is a constant of integration; and since v = — , we may 

multiply by dt and integrate again, obtaining 

(2) x = ±gt 2 + e 1 t+c 2 , 
where c 2 is a second constant of integration. 

To determine the constants of integration, we make use of 
the initial conditions. Suppose the particle is started at the 
point x with the velocity of projection v . Then when t = 0, 
x = x , and v = v . Hence, substituting in (1) and (2), we have 

(10 v o = c v 

(2') x =c 2 . 

Hence the equation of motion is 

x = | gt 2 + v t + x . 



F 

m 


_ m g _ 
m 

dv 


dv 
"dt" 



KINETICS OF A MATERIAL P ARTICLE 95 

The discussion may be made according to the directions given in 
Chapter II. 

47. Resultant force in rectilinear motion. If a particle mov- 
ing in a straight line be acted upon by two or more forces directed 
along the line of motion, the resultant acceleration is the algebraic 
sum of the accelerations due to the separate forces. Suppose the 
particle is acted upon by n forces, F v F 2 , •••, F n . The acceleration 

due to F x isf=^±, to F 2 is/ 2 = ^, •••, to F n isf n = ^, and the 
m m m 

resultant acceleration,/, is given by 

(1) /=/ 1 +/ 2 +-/n= JTl+ ^ + "" J ° - 

m 

Hence, if F denotes the algebraic sum or resultant of the 
collinear forces (F — F x + F 2 + ••• iP„), we have, from (1), 

(2) F = mf 

That is, if a particle moving in a straight line be acted upon by any 
number of forces directed along the line of motion, the product of the 
mass and the acceleration is equal to the resultant force. 

ILLUSTRATIVE EXAMPLES 

1. A heavy body is projected in a vertical direction. Determine the equation 
of motion if the resistance of the air is proportional to the speed. 

Solution. We take the X-axis vertical with positive direction downwards. 
There are two cases : (a) when the body is falling ; (6) when it is rising. 

(a) The weight, acting downwards, is positive and equal to mg. The resistance 
of the air always opposes the motion, and hence, when the body is falling, this 
force is negative. Since the velocity is positive, we have 

Resistance = — fimv, 

where /x is a factor of proportionality. 

The resultant force is F = mg — /xmv. 

(6) When the body is rising, the resistance of the air acts in the same direction 
as the weight, and is, therefore, positive. Since the velocity is negative, we have 

Resistance = — /mv, 

and the resultant force has the same form as in case (a) . 

Hence, in this problem, the force equation is the same when the body is falling 
as when it is rising. 



96 THEOKETICAL MECHANICS 

Since the force is a linear function of v, we use (II) (a), 

d 2 x F n 
dt 2 m 



or 

.,, d 2 x . dx 

dt 2 dt 

The solution of the homogeneous equation (see Calculus, p. 440), 
d' 2 x . dx r. 

is x = Ci + c 2 e - M«. 

We see by inspection that a particular solution of (1) is x — & t, and hence the 
general solution is 

(2) x = & t + os + c 2 e-M«. 

The constants of integration are determined if the initial position, aso, and the 
velocity of projection, v , are known. Differentiating (2), we find the velocity, 

(3) v = 2 — /xc 2 e ~ m*. 

If a; = sco, 17 = ^0, when £ = 0, we find, from (2) and (3), c 2 =-^- — — , 

[A jX 

d = jc + — — — , and hence the equation of motion is 
fj. n 2 

[x n p 2 \/x 2 p } 

2. A box of mass 100 lb. is placed on an elevator which ascends with an acceler- 
ation of 10 ft. per second per second. What pressure does the elevator exert upon 
the box ? 

Solution. Taking the positive direction upwards, and denoting the pressure of 
the elevator on the box by P, we have for the resultant force, 

F = P — mg — mf. 

Substituting the values of m and /, we find 

P= 100(10 + 32) = 4200 poundals. 



PROBLEMS 

1. Find the equation of each of the following rectilinear motions under the 
given conditions : 

(a) F x = mt ; * = 1, v = 0, when t = 1. 

Ans. x = %t* — %t + ±. 

(b) F y = m(t — 1) ; y = 0, v = 1, when t = 0. 

Ans. y = ±t s -lt 2 + t. 

( C ) p x = -^- ; x = 0, v = 0, when t = 2. 

Ans. » = (*-l)[log («-!)-!]. 



KINETICS OF A MATERIAL PARTICLE 97 

(d) F y = m cos t ; ?/ = 0, v = 0, when t = 0. 

^4ws. y = 1 — cos £. 

(e) 2^. = — mx ; x = o, v = 0, when £ = - • 

Ans. x = a sin f. 

(/) jP y = — wiy ; y = 0, v = 1, when £ = tt. 

,4ns. ?/ = — sin f. 

(#) F x = — mx ; x = a cos /3, w = — a sin /S, when £ = 0. 

Ans. x = a cos {t + /3). 

(A) F y =— mk 2 y ; y = a cos /S, v = — ak sin j8, when £ = 0. 

.Arcs. 2/ = « cos (Atf -f j8). 

(£) i^ x = — mn 2 x ; x = an sin v, v = an cos v, when t = 0. 

0") -FV = my ; y = 0, y = 1, when £ = 0. 

^4ns. y = i(e« + e _ <). 

(A;) Fy = —2 my — 2 mu ; y = 0, v = 10, when £ = 0. 

Ans. y = 10 e - ' sink 

(Z) F x = — 25 rax — 6 rat? ; x = a, v = 0, when £ = 0. 

Ans. x = -e- 3 '(4cos4£ + 3sin4£). 

(wi) F x = —2/j.mv — & 2 mx ; x — 0, v = 6, when £ = 0, (A > /*). 

6 



.Arcs, x = , e _ M< sin V& 2 — ix 2 t. 

Vk 2 - ^ 

(n) F x = — 4 mx + 2 w cos £ ; as = 0, v = 0, when t = 0. 

J.ns. x = | (cos t — cos 2 1) . 

(o) F y = — my — msint; y = 0, u = 0, when £ = 0. 

^4jjs. y = \t cos £ — \ sin t 

Q?) F x = — ^mx + ?rc sin rci ; as = 0, v = 0, when t = 0. 

w 1 

^4ws. as = cos At H sin ni. 

A: (k 2 - n 2 ) k 2 - n 2 

(q) Fy = — k 2 my + m cos nt ; y = 0, v = 0, when £ = 0. 

(/•_) i^ = — mx + ?}i sin t -f 3 ra cos 2 £ ; x = 0, v = 0, when £ = 0. 

2. Discuss the following rectilinear motions, taking into account the initial 
conditions. 

(a) f— a 2 + x, given v = c, x = 0, when £ = 0. 

(b) f— x~ 3 ; given v = v , x = x , when £ = £ - 

(c) f=v 2 ; given v = |, x = 0, when t = 0. 

(d) f=av; given u = 6, x = -, when £ =0. 

a 



( e ) /=£- — — ; given ?? = 0, x = 0, when <= 0. 



98 THEOKETICAL MECHANICS 

(/) F x = m(at 2 --) ; given v = -.x=- A , when t = -. 
\ ct/ c 3 c 4 c 

Answers : 

(c) x =- log (1-0. 

(e) ^J_log£±^; U= ^=l^; 

w 2 & & g-kv e kt + e~ kt 

,3 V c*/ c Zc°, c 



+ fl-4c 2 



4 c 4 

3. Show that a particle projected with a velocity v and acted upon by a 
constant force mk will acquire a velocity equal to V2 kx + Vo 2 in moving the 
distance x. 

4. A body is projected vertically upwards with a velocity V. Prove the 
formulas v= V—gt, h= Vt — \gt 2 , where h is the height at any instant. What 

is the greatest height ? A , V 2 

& ° - Ans. h = . 

*g 

5. A body of 25 lb. mass is acted upon by a constant force which in 10 sec. 
gives it a velocity of 75 ft. per second. What is the magnitude of the force in 
poundals ? 

6. A heavy body is projected in a vertical direction. Write the force 
equation and find the equation of motion if the resistance of the air is proportional 
to the square of the speed. 

Ans. When the body is rising, F=mg + fimv 2 ; x = - log sec ( ^/~MJ t + Ci) + c 2 . 

When the body is falling, F=mg — /j.mv 2 ; % =-%/- t + - log (1 — e -2V, ^ +c i) + c 2 . 

7. An elevator, starting from rest, has a downward acceleration of 16 ft. per 
second per second for 1 sec, then moves uniformly for 2 sec, then has an 
upward acceleration of lOf ft. per second per second until it comes to rest. 
(a) How far does it descend ? (5) A person whose weight is 150 lb. experiences 
what pressure from the elevator during each of the three periods of its motion ? 

Ans. (a) 52 ft. (6) 75 lb. ; 150 lb.; 200 lb. 

8. Equal masses of m lb. each rest upon two platforms, one of which has 
at a certain instant a velocity of a ft. per second upwards and the other a velocity 
of b ft. per second downwards. Both platforms have an upward acceleration / 
Compare the pressures of the platforms on the bodies. 

9. A bucket containing 112 lb. of coal is drawn up the shaft of a coal pit 
and the pressure of the coal on the bottom of the bucket is equal to the weight of 
126 lb. Find the acceleration of the bucket. A ns & . 

8 

10. While ascending vertically in a balloon with a velocity v, a man drops a 
stone when h ft. above the ground. Find the time required for the stone to fall to 
the ground. . v + y/v 2 +2gh 



KINETICS OF A MATERIAL PAETICLE 99 

11. A string which can just sustain a mass of 10 lb. against gravity is attached 
to a mass of 2 lb. which rests upon a horizontal table. Supposing that friction 
is _ig. the weight of the body, find the greatest acceleration that can be given to the 
body by means of the string. 

12. A particle moves in a straight line under a force directed towards the origin 

k 2 
and varying inversely as the third power of the distance. Prove v 2 = (- v 2 , if k 2 

x 2 
is the absolute intensity. If the initial distance and velocity are respectively b and 

-,' show that the equation of motion is x' 2 = b 2 — 2 kt. Discuss the motion. 

13. A particle is projected with a velocity v in a medium offering a resistance 
proportional to the square of the velocity. Show that the equation of motion may 

be written s = - log (/xvt + 1). Discuss the motion. 

14. Find the equation of motion if the force is a periodic function of the time. 
Hint. Assume * F x = ma cos kt. Then F x varies from ma to — ma with the 

period — ■ Ans. x = — — cos kt + v t + Xo. 

k k 2 

15. When is the motion in problem 14 periodic and what is the period ? 

2 IT 

Ans. vo = 0, period = — . 
k 

16. A particle describes a straight line under the action of two forces, one con- 
stant and the other an attractive central force proportional to the distance. Show 

that the force equation may be written — ^ = — /x 2 y + /, where /x and / are constants. 
Find the equation of motion and discuss it. 

Ans. y = c cos (fit + v) + -J- , where c and v are constants of integration. 

n 2 

17. Show that the motion in problem 16 is central motion, the center being at 
y = — ^- and attracting directly as the distance, (a) What is the period of the 

/x 2 

motion ? (6) If y = a, v = 0, when t = 0, find the amplitude. Ans. (a) — . 

18. A spring balance is extended | in. by a mass of 1 lb. and the force of the 
spring is proportional to the extension. The spring is then pulled downward and 
released. Show that the force equation has the same form as in problem 16, namely 
d 2 y 

at 2 



= g (1 — 48 y). What is the period of the vibration ? 



Ans. — ^ == = — sec. nearly. 
V48 g n 

* A finite periodic function of the time must have the form A sin {bt + v) or 
A cos (bt+ v), where A, b, and v are constants. 



100 THEORETICAL MECHANICS 

19. A particle is acted upon by a center of force which attracts directly as the 
distance and moves in a medium resisting directly as the velocity. Show that the 

force equation may be written — • + 2 /x (- k 2 x = 0. Find the equation of motion 

., , dt 2 dt 

it /a> k. 



Ans. x = Ae-^cos (Vk 2 — n 2 t + »>), where A and v are constants of integration. 

20. Write the force equation for a particle which is acted upon by an attractive 
center of force proportional to the cube of the distance if the particle moves in a 
medium offering a resistance proportional to the square of the speed. 

21. A central force is attractive and varies as the nth power of the distance. 
If the particle starts from rest at the distance a from the center, find the time of 
arriving at the center when (1) n — 1, (2) n = — 3. 

Ans. (1) T - , (2) -^3, where /x is the absolute intensity. 

2 Vac Vfi 

22. In example 1, p. 95, show that the velocity approaches •? as t increases 

/J- 
indefinitely. Show also that when the particle is projected downwards with this 

limiting velocity, the velocity remains constant, and the motion is uniform. 

48. Curvilinear motion. Axioms on force action. Concurrent 
forces. Three things must be known of a force in order to com- 
pletely determine it, namely, its magnitude, its direction, and its 
point of application. Forces are therefore not vector quantities in 
the sense in which a vector was defined in Chapter III, because the 
line of action of a force cannot be moved without changing the 
effect of the force. We are, however, familiar from experience 
with certain properties of force action which at least suggest 
vector properties. In fact, it is evident that if we confine ourselves 
to forces acting simultaneously upon a material particle, since at any 
instant such forces have the same point of application, magnitude 
and direction are now alone significant. Such forces are said to 
be concurrent. For these forces vector resolution and composition 
have meanings with which we are familiar. These results of ex- 
perience we state in the form of axioms. 

Axiom 1. The acceleration produced by the simultaneous 
action of any number of concurrent forces is equal to the accelera- 
tion which would be produced by their vector resultant. 

In other words, any number of concurrent forces may be re- 
placed by a single force equal to their vector sum. 

Axiom 2. If a force is resolved along any direction, the accel- 
eration due to this component may be found by resolving the original 
acceleration along that direction. 



KINETICS OF A MATERIAL PARTICLE 101 

For example, given a force F which causes at a given instant 
the vector acceleration f in the motion of a material particle of 
mass m. Then, by (I), 

(1) F=mf 

If now we resolve F and f along any directed line Z, the corre- 
sponding components being F : and /,, respectively, then f is the 
acceleration caused by F^ and by (I) and Axiom 2, we shall have 

(2) F,= mf, 

That is, the component of a force along any direction equals the mass 
times the acceleration along that direction. 

49. Curvilinear motion. Suppose a particle moves in a plane 
and is acted upon by n forces, F v F v •••, F n . By the first 
axiom on force action the n forces may be replaced by a single 
resultant force F obtained by the vector addition of the individual 
forces F v F 2 , •••, F n . By the second axiom on force action the 
component of the resultant force F in the direction of the X-axis 
is equal to the mass times the acceleration in the direction of the 
X-axis. Similarly, the component of F in the direction of the Y- 
axis is equal to the mass times the acceleration in the direction of 
the ]T-axis. Hence we have the rectangular force equations for 
plane motion : 



where 



F x = mv x 32* F y = mv y ^ 
doc dy 



F x — sum of ^-components of all forces acting, 
F y = sum of ^/-components of all forces acting. 

The equations of motion are obtained by integrating the force 
equations. 

Equations (III) are the force equations for motion in the XY- 
plane. For motion in space of three dimensions the force equa- 
tions have the same form, the only difference being that the 
additional coordinate z is introduced. See (XII), Art. 37. 

The integration of equations (III) introduces four arbitrary 
constants. Hence to determine the motion completely we must 



102 



THEORETICAL MECHANICS 



have four conditions. These conditions may be the two coordi- 
nates of the initial position (# , y ) and the two components of the 
initial velocity (y^ v y ^). 

From the discussion of Art. 40, we know that 
the resultant acceleration in any curvilinear motion 
is directed towards the concave side of the path 
(in special cases along the tangent). This fact 
enables us to construct in almost all cases the 
beginning of the path. For we may plot the 
initial position (# , y ) and draw the initial velocity (since v x0 
and v y0 are given). Further we may calculate F x and F y for the 
initial conditions and construct the initial force F . The path 
then starts in the direction of v and is concave in the direction of F . 




Po Kvo) 



ILLUSTRATIVE EXAMPLES 

1. Find the equations of motion and the path if F x = 0, F y = mg ; when 

0, x = a, y = 0, v x = b, v y = 0. 

Solution. The force equations are 

d 2 y 



(1) 



at 2 



at 2 



mg. 



Each equation may be integrated separately 
dx „ dy 



(2) 



ci, 



= gt + c 2 . 




dt dt 

A second integration gives 
(3) x = dt + c 3 , y-\ gt 2 + c 2 t + c 4 . 

Substituting the initial conditions in (2) and (3), we find 
(2') b = ci, = c 2 . 

(3') a = c 3 , = c 4 . 

The equations of motion are 

x = bt + a, y=l gt 2 . 
Eliminating t, the equation of the path is found to be 

The path is a parabola with its axis parallel to the Y-axis. 

2. A particle of mass m is acted upon by two forces : (1) one in the direction 
of the T"-axis and equal to mk ; (2) one in a direction making a constant angle a 
with the X-axis and equal to mt 2 . When t = 0, x = 0, y = 0, v x = a, v y = b. Find 
the equations of motion. 



KINETICS OF A MATERIAL PAETICLE 



103 



Solution. The x- and ^-components of the first force are zero and mk, respec- 
tively. The x- and ^-components of the second force are mi 2 cos a and mt' 2 sin a, 
respectively. Hence the force equations are 

(1) m— = mt 2 cos a, m c ^- = mt 2 sin a + mk. 

dt 2 dt 2 



-i^cosa + ci, ^ = 1 t 2 sin a + kt + c 2 . 



Integrating once, 

^ eft ° ' ^ 7 dt 

Integrating a second time, 

(3)' jx = j\t i cosa + c 1 t-\-c 3 ,, 

\y = T \t± sin a + i to 2 + c 2 £ + c 4 . 
Substituting the initial conditions in (2) and (3), we find, 
(2') a = d, & = c 2 , 

(3') = c 3 , = c 4 . 

Hence the equations of motion are 

x = Jj f 4 cos a + «£, ?/ = jL Z 4 sin « + £ to" 2 + &£ . 




PROBLEMS 

1. Find the equations of motion and the path in each of the following :' 

(a) F x = mk, F y = 0; when t = 0, y = 0, x = 0, v x = a, u y = 6. 

^4tts. x = £ to 2 + at, y = bt. Parabola. 

(&) _F X = MX, F y = ; when f = 0, x = 0, y = 0, v x = a, v v = b. 

Ans. x = ^a(e f — e-t), y = bt. 

(c) F x = am, i^ = bm ; when £ = 0, x = 0, ?/ = 0, v x = 0, v y = c. 

J.ns. x = £ a£ 2 , i/ = h bt 2 + c£. Parabola. 

(d) F x = wa, JF'y = my ; when £ = 0, x = 0, y = 0, » x = a, » y = 5. 

Jjis. x = h a{e t — e~t), y = J &(e* — e~<). 

(e) _F X = 0, i^ = my ; when £ = 0, x = 0, y = a, Vx = 1, v y = 0. 

J.ns. x = t, y = %a(e* + e~') . Catenary. 

(/) 2P X= mx, i^ y = my ; when £ = 0, x = a, y = 0, v x = 0, v y = b. 

Ans. x = J a(e' + e - ')? 2/ — I ^( e * — e_ 0- Hyperbola. 

(g) F x = 0, F y = mvy ; when £ = 0, -x = 0, y = 1, fl x = 1, % = 1. 

J.ws. x = t, y = e*. Curve y = e x . 

(h) F x - — mv x 2 , F y =0 ; when £ = 0, x = 0, y = 0, v x = 1, v y = 1. 

^4ns. x = log (t -\-Y), y — t. Carve x = log (1 + y). 

(i) F x = — m sin t, J^ = mv y 2 ; when £ = 0, x = 0, y = 0, tfc = 1, v y = 1. 

^Ins. x= sin c, ?/ = logf )• 

( j) F x = -, F y = 0; when J = 0, x = 9, y = 9, r* = 3, v y = 2. 



Ans. x = ( 2t + 9)2 , y = 2 J + 9. Curve 9 x 2 = y\ 



104 THEORETICAL MECHANICS 

(k) F x = 0, F y = mv ; when i = 0, x = 0, y = 1, v x = 1, v y = 0. 

-4ws. x = t, y = e —tl — Catenary. 
z 

(I) F x = ma, F y = w6, i^ = mc ; when £ = 0, x = 0, */ = 1, 2 = 2, v x = 0, 

V y = 0, t>„ = 1. Ans x _ i ^ y = ^ w + ^ g = j ^ + f + 2 _ 

Parabola in plane ay = bx + a. 
(to) 2^ = 0, i^ = 0, F z = 0. ^4ns. Straight line. 

(w) F x =— mx, F y = — my, F z — 0. 

Ans. Helix if initial position is (a, 0, 0) and velocity (0, b, c). 

2. Show that the path is necessarily a straight line or a parabola if the force 
is constant. 

3. A particle is acted upon by two forces : (1) one parallel to the X-axis and 
equal to m(t— 1) ; (2) one in a direction making an angle of 30° with the X-axis 
and equal to to sin t. When t = 0, x = 0, y — a, v x = 6, v y = 0. . Find the equations 
of motion. 

4. A particle is acted upon by two forces : (1) one parallel to the X-axis and 
equal to — mx ; (2) one in a direction making an angle of 135° with the X-axis and 
equal to mk. When t = 0, x = a, y = 0, v x — 0, v y = 0. Find the equations of 
motion. 

5. A particle is acted upon by three forces : (1) one parallel to the F-axis and 
equal to — mk' 2 y ; (2) one parallel to the X-axis and equal to m(t' 2 — t); (3) one in 
a direction making an angle of 210° with the X-axis and equal to mk 2 . When t = 0, 
x = 0, y = 0, v x = 0, v y = 0. Find the equations of motion. 

6. A particle of mass 12 lb. is moving in a northeast direction with a 
speed of 6 ft. per second. It is acted upon by two forces, one of 48 poundals 
towards the north, the other of 72 poundals towards the east. Find its position after 
the lapse of one second. 

7. A particle of mass 10 lb. is moving towards the north with a speed of 20 
ft. per second. It is acted upon by three constant forces : (1) 10 poundals towards 
the northeast, (2) 20 poundals towards the east, (3) 15 poundals towards the south. 
Find its position and the components of its velocity after the lapse of 3 sec. 

8. A particle of mass to free to move in the XF-plane is subject to a force 
whose axial components are F x = — 16 tox, F y = — 4 my. The initial conditions are 
x = 1, v x =0, y = 0, v y = 2, when t = 0. Find the equations of motion and the 
equation of the path. Discuss the motion. 

9. The axial components of the force causing a plane curvilinear motion are 
F x = — mx, F y — — 4 my. The initial conditions are x = 0, y = 1, v x = 1, v y = 0, 
when t = 0. Derive the equations of motion, discuss them, and draw the path. 

10. A particle of mass m moves in the XT-plane under the action of a force 
whose axial components are F x = — mx, F y = — my. The initial conditions are 
x = 2 a, y - 0, v x = 0, v y = a, when t = 0. Derive the equations of motion. Dis- 
cuss the motion. 

11. Discuss the motion of the particle in problem 10 if the initial conditions are 
x — 0, v x = a, y = a, v u = 0, when t = 0. 



KINETICS OF A MATERIAL PARTICLE 105 

12. A particle of unit mass moves in the XF-plane under the action of a force 
which is directed always towards the origin, and its magnitude is proportional to the 
distance of the particle from the origin, (a) Denoting the magnitude of the force 
when the particle is at unit distance by k 2 , find and discuss the equations of motion 
if the initial conditions are x= a, y= 0, v x = 0, v y = kb, when t— 0. (ft) Prove 
that for any initial conditions the path is an ellipse with center at the origin, 
(c) Under what initial conditions can the ellipse degenerate into a straight line ? 

13. A particle of unit mass moves in the XT-plane under the action of a repul- 
sive force from the origin. The magnitude of the force is proportional to the distance 
from the origin, (a) Find and discuss the equations of motion if the initial condi- 
tions are x = a, y = 0, v x = 0, v y = ft, when t= 0. (ft) Prove that, for any initial 
conditions, the path is an hyperbola with center at the origin, (c) Under what 
initial conditions can the hyperbola degenerate into a straight line ? 

50. Intrinsic force equations. In Art. 40, the tangential and 
normal components of the acceleration were found. If, then, the 
resultant force F, producing a motion in the plane, is resolved at 
any instant along tangent and normal, the corresponding com- 
ponents being F t and _F re , we shall have 

( iv) Ft = mft = m f t = mv f s , *, = ™/„ = ?f . 

These equations, being entirely independent of coordinates, are 
called the intrinsic force equations. 

The component F n is directed always towards 
the center of curvature. Motion along any plane 
curve may therefore be said to be caused by the 
simultaneous action of a tangential and a nor- 
mal force. The change of motion involved, that 
is the change in the vector velocity, may be 
explained thus. The resultant tangential force 
F t changes only the speed, that is the magnitude of the vector 
velocity. The resultant normal force causes change in direction 
of the velocity. If the speed is constant, the resultant tangential 
force is zero. If the path is rectilinear, the resultant normal 
force is zero. 

51. Polar equations. In many problems the use of polar co- 
ordinates is advantageous. If all forces causing a motion are re- 
solved along and perpendicular to the radius vector drawn to any 
point of the path, we shall have, by Art. 41, 




106 



THEOKETICAL MECHANICS 



(1) 



F p = mf p = m 

Fa 



d?p _ (dd\ 2 



dt 2 



P[ dtJ 



/. m d 



2 dF\ 



When F p and F e are given, the equations (1) become simultaneous 
differential equations of the second order, from the integration of 
which p and 6 are to be determined as functions of t. 



By introducing into (1) the radial velocity 



dp 



dt 



and 



the angular velocity ( a> 
(V) 



d0 



dt 



about the origin, we obtain 



m-?»* 



^=7l(p 2 »)- 



Illustrative Example. A particle moves under the action of a force directed 
always along the radius vector, the magnitude of the force being inversely pro- 
portional to the cube of the distance. Discuss the motion. 



Solution. Since the force is radial, we have . 

(1) 




F p =^* F d 
P s 



Hence from the force equations (V) , 



(2) 



dV r 



P" 2 = ^1 



d 



(p 2 co) = 0. 



dt p 8 ' dt 

The second of these gives, by integration, 



(3) 



p 2 u) = constant, or w = — 



Substituting in the first, we get after reducing, since v p = 



(4) 



d?P = _k 
dt 2 p3 



where k = — p. 



dp 



Multiplying both members by 2 ^ dt, and integrating, 



(5) 



dpV 

dt) 



Cl + 



To obtain a simple solution,! assume as one initial condition, when p = a, 

y/k 
^p = Then c\ = 0, and (5) becomes 



* When p = 1 and m = 1, the force = /«.. Hence ^ is the magnitude of the force upon 
unit mass at unit distance. It is convenient to call n the absolute intensity of the force, 
t For the general case see problem 5 below. 



KINETICS OF A MATERIAL PARTICLE 107 

(6) d -P=^,ovdt = ^P. 
dt p y/k 

Evidently k must be positive, that is, since k = — p. — c 2 , p. must be negative, 
and from (2), the force is now an attraction. 

The equation of the path is found directly from (3) and (6) . For, 

(7) » = *=± % ...d» = ^d* = -5L*. 

dt /> 2 ' p 2 V k P 



Integrating, we have, 



= -^logp + c' 2 , 
Vk 



or also —d = log op, or p = be — 0, 

c c 

by changing the form of the constant of integration. The path is therefore a 
logarithmic spiral. 

The assumed initial radial velocity lv p = — - ] may be interpreted thus. In- 
tegrating (4) with the assumption that v p = when p =oo , we find C\ = 0, and 

.*. v 2 = — . If p — a, this gives v p = ■ — -', the value assumed above. That is, the 
P 2 a 

initial radial velocity is assumed to be that acquired in moving in from infinity. 
To find the resultant initial velocity, we have from (7) for initial angular ve- 
locity, o> = —. .-. v e = pw = aw = - , and v 2 = v 2 + v e 2 = —+ — = — -£- , since 
a 2 a p a 2 a 2 a 2 

k — — p. — c 2 . Hence r = — :: ^', the initial velocity. 
a 

The discussion leads to' the 

Theorem. Given an attractive central force varying inversely as the cube of the 
distance. Under such a force a particle will describe a logarithmic spiral if pro- 
jected in any direction with a velocity equal to — -, where a is the distance from the 

a 
center and p. the absolute intensity of the force. 



PROBLEMS 

1. Show that a particle can move freely in a conic section with focus at the 
origin when acted upon by an attractive center of force varying as the inverse 
square of the distance. 

2. Show that a particle can move freely in a conic section with center at the 
origin when acted upon by a center of force varying directly as the distance. 

3. Show that a particle can move freely in a circle when acted upon by a con- 
stant attractive central force. 

4. Show that a particle can move ireely in a circle under a center of force di- 
rected towards a point on the circumference. Show that the law of the force is the 
inverse fifth power of the distance. 



108 THEOKETICAL MECHANICS 

5. Find the general path under a central force varying inversely as the cube of 
the distance. 

Hint. Integrate equation (5) of the Illustrative Example, p. 106, without 
specializing c±. 

Ans. k < 0, - = a cos (~0 + p\ ; k = 0, pd = a; 

k > 0, - = c\e c + Coe c . 
P » 

6. A particle is acted upon by two centers of force, one attractive and the 
other repulsive, but of equal magnitude at any point. Show that the path is a 
parabola. 

7 . A particle is acted upon by a central force proportional to the distance and 
a constant force. Show that the path is a conic section. 




CHAPTER V 

WORK, ENERGY, IMPULSE 

52. Work. A force is said to do work when its point of appli- 
cation undergoes a displacement. The amount of work done by 
a constant force is equal to the product of the force and the dis- 
placement in the direction of the force. For example, consider 
the force of gravity. If a particle of mass m drops vertically 
through a distance of h units, the work done by 
the force of gravity is mgh units. If the parti- 
cle rises vertically a distance of h units, the 
work done by gravity is — mgh units.* If the 
particle slides a distance of s units down a smooth plane whose 
inclination is a, the work done by the force of gravity is 
mgs sin a units, or mgh units, where h = s sin a is the vertical dis- 
tance moved through, that is, h is the distance in the direction of 
the force. If a = 0, that is, if the particle slides along a hori- 
zontal plane, the work done by the force of gravity is zero. 

Suppose the particle moves along the straight line OX under 
the action of a force directed always along the line, but whose 
j> magnitude is variable. The 

O a dx b X law of the variation of the force 

being known, its value is known for any position of the particle 
on the line. The small amount of work dW,f done by the 
force F, while the particle moves a small distance dx from the 
point Pj, is approximately equal to F x dx, where F x is the value 
of the force at the point P 1 (x = x-^). Applying the princi- 
ples of the Calculus, Chapter XXX, it is evident that the total 
work done by the force F while the particle moves from the point 
x = a to the point x = b, is the definite integral of F with respect 
to x from a to b ; that is, 

(1) W= f'Fdx. 

* When the work done by a force is negative, it is sometimes said that the particle 
does work against the force. 

t dW is called the element of work ; dx, the element of distance. 

109 



110 THEORETICAL MECHANICS 

Suppose the particle moves in a plane under the action of a 
resultant force which is variable in magnitude and direction. 

Let P x represent any position of the par- 
ticle on the path AB. Let P X C repre- 
sent the resultant force in magnitude 
and direction. The element of work 
dW corresponding to an element of 
arc ds is equal approximately to the 

value of the force F at P x multiplied 

x by ds cos 0, where 6 is the angle between 

the direction of the force and the tangent to the curve ; that is, 

(2) dW= F cos Ods. 

But F cos 6 = F t where F t is the tangential component of the 
force F. 

The total work done by the force F while the particle moves 
from an initial position s on the path to any other position s is 
obtained by integration. 




(3) W=J S F t 



But by (II), Art. 32, and (4), Art. 40, we have 

_. _. dx -r, dy 

F t = F x -y-+F v -?-. 
1 x ds y ds 

By substitution (3) becomes the work integral : 

(I) W=C ,V (JF X doc + F y dy) . 

An important consequence of formula (I) is the application 
to plane motion under a constant force. Let the direction of the 
P"-axis agree with the direction of the force. Then F x = 0, F y = F, 
and formula (I) becomes 






Hence in any plane motion the work done by a constant force is 
equal to the product of the force and the distance moved through 
in the direction of the force. The work is independent of the path. 
For example, if a particle moves on any plane curve the work 
done by the force of gravity is equal to mgh where h represents 
the vertical distance moved through. 



WORK, ENERGY, IMPULSE 



111 



Dimensions. Since work has been denned as the product of 
force by distance, the relation between the derived unit of work 
and the fundamental units of mass, distance, and time is given by 
the dimensional equation : 

Work = mass x length 2 
time 2 

ILLUSTRATIVE EXAMPLES 

1. A unit particle describes the parabola y 2 — 4x from x = 1 to x = 4. Find 
the work done by the force whose axial components are F x — 2 my, F y =0. 

Solution. Since m = 1, (I) becomes 

W= [ X= \ydx = 4 C Vx dx = f [^j 4 _ §e units# 





2. Find the work done by a force whose axial components are 

(1) F x = — mky, F y = mkx, 
upon a particle moving along the parabola 

(2) x' i = 4 + y, 

from x = to x = 2. 

Solution. The work integral (I) may be obtained as a function of x in the fol- 
lowing manner. From (2), y =x 2 —4, .•. dy = 2xdx, and (I) becomes 

W=C (— mk (x 2 -4) dx+ mkx • 2xdx) = mkC (x 2 + 4) dx = lOf mk. 



PROBLEMS 

1. Show that in polar coordinates work is given by the formula 



W = C ,e AF P d P + F eP dd). 



2. A body whose mass is m falls vertically to the earth's surface from a height 
equal to the radius B. Compute the work done by the earth's attraction during 
the fall. Ans. \ mgB units. 



112 THEORETICAL MECHANICS 

3. A body is suspended by an elastic string whose unstretched length is 
4 ft. Under a pull * of 100 poundals the string stretches to a length of 5 ft. 
Required the work done on the body by the tension of the string while its length 
changes from 6 It. to 4 ft. Ans. 200 units. 

4. A unit particle describes the circle x 2 + y 2 = a 2 , from (0, a) to (a, 0). 

Find the work done by the force whose axial components are F x = ky 2 , F y = kxy. 

k 
Ans. - a 3 units. 
* 3 

5. A particle describes the circle x 2 + y 2 = a 2 from the point (0, a) to the 
point (a, 0). Prove that the work done by the force whose axial components are 
F x = mxy, F y = my 2 is zero. 

6. Calculate the work in problem 5 if the particle moves from the point 
(0, — a) to the point (a, 0). 

7. A particle describes the circle x 2 + y 2 = a 2 from (0, a) to (a, 0). Find the 
work done by the force whose axial components are F x = my, F y = my 2 . 

Ans. ma 2 (^-^\. 

8. A unit particle describes the curve x = & — e~y from x = to x = 2. Find 
the work done by the force whose axial components are F x = mx, F y = 0. 





















Am 


?. 2 units. 




9. A particle of 


mass 


m describes the 


hyper 1 


bola 


X 2 

a 5 " 


y 2 
b 2 ~ 


-. 1 from 


the point 


(a, 


0) to the point (ot 


"o» Vo)- 


Find the 


work done 


by 


the force 


whose axial com- 


poi 


lents are F x = mx, 


Fy = 


my. 






Ans. 


| ! « 


2 + 2/o 2 - 


a 2 ) units. 



10. The equations of motion of a unit particle are x = t, y = eK Compute the 
work done by the resultant force during the time from t = to t = 3. 

Ans. I (e 6 — 1) units. 

11. The equations of motion of a particle of mass m are x = log (t -f 1), y — t. 
Compute the work done by the resultant force during the time from t = to t = 3. 

Ans. i| units. 

ill 

12. A particle describes the parabola x 2 + y 2 — a 2 from (0, a) to (a, 0). Find 

the work done by the force whose axial components are F x = kmy, F y = kmxy. 

Ans. ^^ (5 -a) units. 
30 v J 



X" V 

13. A unit particle describes the ellipse — + ^ = 1 under the action of a force 

whose axial components are F x = — x, F y = — y. Compute the work done while 
the particle moves from (0, 6) to (a, 0) . 

14. Compute the work done when a particle describes one half of the ellipse 

9 = : — -, under the action of a force directed always towards the origin and 

1 — e cos0 J ° 

varying inversely as the square of the distance. 

*The law of force for an elastic string is Hooke's Law. The tension of an elas- 
tic string is proportional to the extension. 



WORK, ENERGY, IMPULSE 113 

53. Kinetic energy. A particle is said to possess energy when 
its condition is such that it can do work against a force which 
may be applied to it. If a particle, acted upon by no forces, has 
a velocity v, it will continue to move uniformly in a straight line. 
Suppose a force, acting upon the particle, brings it to rest after 
moving through a certain distance. The force does work upon the 
particle, and the particle is said to do work against the force. A 
particle in motion, therefore, possesses energy called kinetic energy. 
The kinetic energy of a particle of mass m, moving with velocity 
v, is defined as one half the product of the mass by the square of the 
speed. That is, 

Kinetic energy = J mv 2 . 

Dimensions. From the preceding definition it follows that the 
derived unit of kinetic energy is expressed in terms of the funda- 
mental units of mass, length, and time by the dimensional equation 

ir . , • mass x length 2 

Kinetic energy = — ^ 

time 2 

Comparison with Art. 52 shows that the dimensions of kinetic 
energy are the same a*s the dimensions of work. 

If a particle under the action of a resultant force F moves 
along the X-axis from the initial position x to any other position 
#, the work done is given by 

(1) w=f 



2 Fdx. 
Now F= m»~, and by substitution, (1) becomes 

CLX 



(2) W= f 



mvdv. 



If v is the velocity at the point x and v the velocity at the. 
point z, we obtain by integration of (2), 



(3) Tf= | ( „2_ t , o 2 ) 



2\ _ 1 



mv c — Amv f 



The work is therefore expressed in terms of the kinetic energy 
at the final and initial positions of the particle. If the initial 
velocity is zero, (3) becomes 

W —\mv 1 ^ 

and we have proved for rectilinear motion the 



114 



THEORETICAL MECHANICS 



Theorem. The work done by the resultant force while a particle 
starting from rest acquires the velocity v is equal to the kinetic energy 
of the particle. 

If the velocity of a particle changes from the initial value v Q 
to the final value v, equation (3) gives for rectilinear motion the 

Theorem. In the displacement of a particle the work done by 
the resultant force is equal to the difference of the final kinetic energy 
and the initial kinetic energy. 

Comparison of equations (1) and (3) gives 
the energy equation 

m 




en) 



(v 2 - W) 



Jx 



Fdx, 



for rectilinear motion. When the force F is a 
given function of the distance, integration of (II) 
leads to an expression for the speed as a function 
of the distance. 

Illustrative Example. Find the velocity of a body fall- 
ing from a distance h from the center of the earth. 
Solution. The earth's attraction varies inversely as the square of the distance 
from the center. Hence at P where OP = x, we have F x — — ^— , where k 2 is the 



absolute intensity. Using the energy equation, the result is 
,2 — f x -n j C x nik 2 



i mv 2 = C F x dx = ( x -™^dx = mk 2 

Jh Jh x 2 



U hj' 



since 

But • 

Hence 

(1) 



x 



OA = h, vo = 0, when t 



~ 2/l_l 



x h 

— mg when x = radius of earth = B. 

. mk 2 
B 2 



v 2 = 2 gB 2 



— mg, and k 2 — B 2 g. 
1 1 



When the body reaches the surface, x = B, and (1) becomes 

2gB 2 



2gB 



h 



If the particle falls from an infinite distance (h = oo ), the velocity upon reach- 
ing the surface of the earth is V2 gB. Expressing B and g in miles, we have 
, and the velocity from infinity is approximately 7 miles per 



B = 4000, 
second. 



32 
5280 



WORK, ENERGY, IMPULSE 115 

If the particle moves along a plane curve from the initial posi- 
tion s to any other position 8, where s denotes the length of arc 
measured from a fixed point on the curve, the work done upon it 
by a force F is given by (3), Art. 52 : 



(4) W= f S F t ds. 



Now, by (IV), Art. 50, F t =mv-^, and by substitution, (4) becomes 

as 



(5) W=f 



mvdv. 



If v is the velocity at the point s , and v the velocity at the 
point s, we obtain by integration of (5) 



m 



(6) W= — (V — v 2 ) = J mv 2 — J mv 



Equation (6) shows that the preceding theorems, p. 114, hold 
also for plane curvilinear motion. 

Comparison of (6) and (I) gives the energy equation for curvi- 
linear motion : 



(III) 5f«» - vA = C\F x dx + F y dy). 



When the components F x and F y are given functions of the 
position (#, y), integration of (III) leads to an expression for the 
speed as a function of the position. 

Illustrative Example. A bullet with an initial velocity of 1500 ft. per 
second, strikes a target at 1200 yd. distance with a velocity of 900 ft. per second. 
Supposing the range of the bullet is horizontal, compare the mean resistance of the 
air with the weight of the bullet. 

Solution. Denoting the constant resistance of the air by — F, we find that the 
work done by this force is — 3600 F. Hence the work equation gives 



3600 F=™(& - vo 2 ) = -[900 2 - 1500^ 



whence 


F= 200 m. 


Since 


the weight = 32 w, 


we have 


resistance _ 25 
weight 4 



116 THEORETICAL MECHANICS 

PROBLEMS 

1. A particle is projected in any direction with the velocity u and then falls 
freely under the action of gravity. Find the energy equation. 

Ans. v 2 —v 2 = 2 gh, where h = vertical distance. 

2. A particle moves in a circle of radius a under a constant tangential force mf. 

Find the energy equation. v -2 _ v 2 

Ans. adf = - fSL (0 = angle moved through). 

3. Find the energy equation for a force parallel to one axis and proportional 
to the distance from the other. 

4. With what velocity must a particle be projected from the surface of the 
earth in order that it may never return, no force except the earth's attraction being 
supposed to act ? 

5. A bullet moving with the speed of 1000 ft. per second has its speed reduced 
by 100 ft. per second in passing through a plank. How many such planks would 
the bullet penetrate ? Ans. 5- 5 -. 

6. A bullet fired with a velocity of 1000 ft. per second penetrates a block of 
wood to a depth of 12 in. Assuming the resistance of the wood to be constant, 
prove that if fired through a board 2 in. thick, its velocity would be reduced by 
about 87 ft. per second. 

7. A laborer has to send bricks to a bricklayer at a height of 10 ft. He 
throws them up so that they reach the bricklayer with a velocity of 10 ft. per second. 
What proportion of his work could he save if he threw them so as only just to reach 
the bricklayer ? 

8. A particle moves under a central attraction proportional to the distance. 
Find the energy equation. Ans . ^2 _ p ^ =v i_ ^2. 

9. A particle moves under a central attraction inversely proportional to the 

square of the distance. Find the energy equation. 2 A* 2 k 

Ans. = v 2 — vq 2 . 

P Po 

54. Constrained motion. Dynamic pressure. The motion of 
a particle is said to be constrained when it is confined to a certain 
curve or surface ; for example, a bead sliding on a wire, or swing- 
ing on a string, or moving on an inclined plane. In constrained 
motion, the forces acting upon a particle may be divided into 
two classes:* (1) the impressed forces; and (2) the pressure of 
the constraint. 

Two cases are to be distinguished. (1) On a smooth curve 
the tangential component of the force of constraint is zero, that 

* The difference between the impressed forces and the force of constraint is that 
the former are given directly, while the latter is not given directly, but its effect upon 
the motion is specified. 



WORK, ENERGY, IMPULSE 117 

is, there is no friction. (2) On a rough curve the tangential com- 
ponent of the force of constraint is called the sliding friction. 
We suppose, for the present, that the curve is smooth. The case 
of a rough inclined plane is treated later, Art. 67. 

Making use of the intrinsic force equations, Art. 50, we have 

■W v f normal 1 (normal 
C 1 V) ~W = Fn= ™P ressed \ + 1 pressure 

{ force J [ 

The impressed forces being known, their normal components are 
known ; _R, the radius of curvature, may be calculated from the 
equation of the path, and v 2 may be found from the energy equa- 
tion. Hence formula (IV) determines the normal pressure. 

The normal pressure is exerted by the path upon the particle. 
In many practical problems it is important to know the normal 
pressure exerted by the particle upon the path. This is given by 

Newton's Third Law of Motion. To every action there 
is a corresponding reaction, equal in magnitude but opposite in 
direction. 

The curve exerts a normal pressure on the particle. Hence 
the particle exerts a pressure equal in magnitude but opposite in 
direction on the path. , In the case of a bead sliding on a surface, 
this is called the dynamic pressure. In the case of a particle 
swinging on a string, it is the tension in the string. From (IV) 
we obtain 

f dynamic 1 f normal ) 



,2 



(V) { pressure [■ = j impressed [ —^~. 

{ on path J { force J 

The motion is said to be free when the dynamic pressure is 
zero. 

It must be remembered that normal forces are resolved along 1 the 

mv* 
~E 
center of curvature or inwards. If the nor- 
mal impressed force acts inwards also, as in 
(a), then the dynamic pressure equals numeri- «v (b) 

cally the difference of the other two forces. On the other hand, 
if the normal impressed force acts outwards, as in (6), then the 
dynamic pressure equals numerically the sum of the other forces. 



directed normal. The resultant force, — — , acts always towards the, 



y x 



118 THEORETICAL MECHANICS 

When v = 0, that is, when the particle is at rest, the pressure 
on the path equals the normal impressed force. For this reason, 
in this case, the latter is also called the static pressure. The term 

— , which gives the change in pressure due to the motion, is 

commonly called centrifugal force. From (IV) this force is equal 
and opposite to the resultant normal force and acts always out- 
wards.* 

In terms of static pressure and centrifugal force,- equation (V) 
may be written 

f dynamic 1 __ J static 1 _i_ f centrifugal 1 
I pressure J 1 pressure J 1 force J 

Since the centrifugal force acts always outwards,. it follows that 
the dynamic pressure is numerically equal to the sum or difference 
of the centrifugal force and static pressure according as the latter 
is outwards or inwards. 

ILLUSTRATIVE EXAMPLES 

1. A heavy particle is constrained to move in a smooth fixed semicircle whose 
plane is vertical. Find the pressure at the lowest point. 

Solution. The impressed force is weight. If the particle falls from P to A, the 
work done is mg • AM — mg (AC— MC) = mga (1 — cos a), if a is the radius. 

Hence, using the energy equation, and assuming the 
r~ particle to start from rest at P, we have 



\ / The normal impressed force acting outwards, we ob- 
"""xTrp tain for the pressure at the lowest point, 

t m g -^T^L — m g + 2 mg(l — cos a) = mg(S — 2cos«). 

m 9 a 

If the particle starts at the highest point of the semicircle, a = — , then the pressure 

equals 3 mg. That is, the pressure at the lowest point is trebled by the motion. This 
increase of the static pressure by motion is a matter of importance. For example, 
in a scenic railway the structure must withstand not only the weight of a car and 
its occupants, but also the added pressure due to motion. This added pressure 
equals the centrifugal force. 

* It must be clearly understood that the centrifugal force is not an actual force act- 
ing on the particle. It is the reaction of the particle against the normal component of 
the resultant force. By the first law of motion the particle tends to move in a straight 
line. If it moves in a curved path, centrifugal force is a convenient term to designate 
the magnitude of the normal force which must act on a particle of mass m and velocity 

v, in order to produce the curvature — in the path. 

R 



WORK, ENERGY, IMPULSE 



119 



2. A heavy particle is suspended from a fixed point by a string of length a. A 
horizontal velocity v is suddenly imparted to the particle so that it begins to de- 
scribe a vertical circle. Determine whether the particle will oscillate or the string 
become slack. 

Solution. The work done by the impressed 
force weight when the particle moves from to P 
is negative and equal to — mgy, if OM=y. Here 
the energy equation gives 



(2) 



j 1 mv 2 — \ mva 2 

\ V 2= Vo 2->2 



, or 



%gy- 



The normal impressed force is the component of 
the weight along the radius and equals 



mg cos a= mg 



MC _ a-y 
CP a 



mg. 



Hence from (V), since the dynamic pressure is in 
this case the tension, we get 

,2 




Tension 



mg a- 1 + mv L = mg 
a a 



M-C^f) 



(3) 



mg + 



3 mgy 



If the particle oscillates, the velocity at the highest point must vanish. From 

v 2 
(2), if v 2 — 2 gy = 0, we obtain y = -S- as the corresponding height. Since this 

2 g 

height must be less than a radius, a necessary condition for oscillation is vo 2 < 2 ag. 

If the string becomes slack, the tension must vanish. From (3), if 

, mv,? 3 may n 
mg -\ 2 ^- = 0, 



we have y = r ° + a 9 as the corresponding height. This must be less than 2 a, 

3g 

and from Vn + a ® < 2 a, we get v 2 < 5 ag as a necessary condition for the tension 
vanishing. 

Subtracting the two heights found, we obtain 



(4) 



^o 2 + ag % 2 
3a 2a 



2ag 



vg 



Comparison of the inequalities and interpretation of (4) gives the criterion : 
If v 2 > 5 ag, the particle describes the whole circle. If v 2 < 5 ag and > 2 ag, the 
tension vanishes, the string becomes slack and the particle will leave the circle and 
fall freely. If v 2 < 2 ag, the equation (4) shows that the velocity vanishes before 
the tension, hence the particle will oscillate. 



3. A particle is attached to the end of a fine thread which just winds around 
the circumference of a circle of radius a at whose center there is a repulsive force 
varying as the distance and of absolute intensity /x. Find the time of unwinding 
and the tension at any time. 



120 THEORETICAL MECHANICS 

Solution. The path is an involute of the given circle. Since the impressed 
force acts along the radius vector OP, we have, using polar coordinates, 

F p = raw, F e = 0. 
Hence, by the energy equation, we have 

i mv 2 = I mixpdp = \ mp (p 2 — a 2 ) , 

%J a 

since the particle is at rest at A where p = a. 

(1) .-. v 2 = p(p 2 -a 2 ). 

Now introduce properties of the involute. PS is the normal at P, .*. \p = angle 




SOP. Also PS = arc SA. Hence cos yp = cos SOP = ^- = ~. PS = act. Since 

o 2 o OP p 

also OP 2 = OS 2 + £P 2 , we have 

Hence (1) becomes 

(2) v 2 = p (p 2 - a 2 ) = pa 2 a 2 . 

Now v p = v cos \f/ = v - . .'. using (2), 



(3) v p ~*l>-v*=^ aV l*-<* , or -M?— = v Ws, 

dt p p V/3 2 — a 2 

Integrating, remembering that p = a when £ = 0, we obtain 

(4) Vp 2 - a 2 = Vp at, or also, a = \ r /xt, [by (2)]. 

2 7T 

When the string is unwound, a = 2 tt, and hence £ = — -. 

Vp 



WORK, ENERGY, IMPULSE 121 

The pressure on the path is the tension. In the figure, 

F n = F sin V = F sin POS = F~= mpp — = m/iaa. 

Also ^- = m ^ a ~ a " — mfiaa. Since F n acts outwards, the tension is the sum of F n 
B SP 3 

and the centrifugal force ; that is, tension = 2 m/xact = 2 mix 2 at. Hence the tension 

constantly increases and reaches the maximum value 4 nifiair. 

PROBLEMS 

1. A heavy particle slides from rest down a smooth curve. If h is the vertical 
height fallen through, prove v 2 = 2 gh. 

2. A heavy particle falls down an inclined plane whose inclination to the 
horizon is a. Show that the dynamic pressure is constant and equal to mg cos a. 

3. A particle slides down a smooth plane whose inclination to the horizon is 
30°. What is the velocity when it has traversed a distance of 20 ft. ? 

Ans. 25.3 ft. per second. 

4. A heavy particle slides on the exterior of a vertical circle. If the particle 
is just started at the highest point, show that it will leave the circle and fall freely 
after sliding through a vertical height equal to one third of the radius. 

5. A heavy particle is constrained to move in a circle under a repulsive center 
of force lying on the circumference and varying as the distance. The particle just 
starts from rest at the center of force. Find the pressure. 

Ans. — iR££^ where p = distance from center of force and a = radius. 
2a 

6. A heavy bead is constrained to slide on a smooth wire of the shape given 
by one of the following equations (assuming the y-axis vertically upwards). It 
starts at the point indicated. Find the pressure at the end point given. 

(a) y 2 = 4z, start (4, 4), end (0,0). 

(6) 4x 2 + y 2 = l6, 

(c) y = cos x, 

(d) x — a arc vers ^ — y/2 ay — y' 2 , 

a 

(e) y = x s , 

(/) y = - x 2 , 

7. If, in problem 0, the bead slides on the exterior of the given curve, find 
where it will leave the curve. 

8. A heavy bead is constrained to move on a smooth vertical circular wire. 
Show that the bead will describe the whole circle if projected from the lowest point 
with a velocity greater than 2Vag, where a is the radius of the wire. 

9. A particle is attached by a string of length I to a point in the same hori- 
zontal plane. What is the least upward velocity with which it must be projected 
so that it shall describe a circle ? Ans. V3 gl. 



(0,4), < 


(2,0). 


(0,1), ' 


64 


Or, 2 a), ' 


(2tt,0). 


(2,8), « 


(0,0). 


(0, 0), ' 


' (1,-1). 



122 THEORETICAL MECHANICS 

10. A particle hangs freely from a string of length I ; it is projected horizon- 
tally with a velocity a/4 Ig. Find how high it rises before the string. becomes slack. 

Ans. f I. 

11. A weight of 10 lb. is fastened by a string which passes through a hole in a 
smooth horizontal table to a weight of 1 lb. which hangs vertically. The first 
weight is revolving on a table about the hole as center. How many revolutions are 
there per minute if the horizontal portion of the string is 15 in. long? ' 

12. A ball is hung by a string in a railway carriage which is rounding a curve 
of 1000 ft. radius with a velocity of 30 mi. an hour. Find the inclination of the 
string to the vertical. Ans. arc tan aWo* 

13. A heavy particle is suspended by a string from a fixed point and rotates 
in a vertical circle. Show that the sum of the tensions of the string when the par- 
ticle is at opposite ends of a diameter is the same for all diameters. 

14. A particle falls down a vertical circle, starting from rest at the highest 
point. If, when at any point, its velocity be resolved into two components, — one 
passing through the center, the other through the lowest point of the circle, — prove 
that the latter is of constant magnitude. 

15. A bead is constrained to move on a circular wire and is acted upon by a 
central force tending to a point on the circumference and varying inversely as the 
fifth power of the distance. Show that the pressure is constant. 

16. A body describes a parabola under the action of two equal forces, one 
tending to the focus and varying inversely as the distance, the other parallel to the 
axis. Show that the speed is constant. 

17. A particle is constrained to move on a logarithmic spiral p=ae mQ in a 
central field for which the force varies inversely as the cube of the distance. Show 
that the pressure varies inversely as the distance. When is the motion free ? 

18. A particle describes a parabola freely under the action of two forces, one 
a repulsion from the focus varying as the distance, the other parallel to the axis 
and equal in magnitude to three times the former. Show that the initial velocity is 
2 p V/a, where p is the distance from the focus and fx is the absolute intensity of the 
repulsive force. 



55. Units of work and energy. Power. Work is the product 
of force and displacement. Hence unit work is done when unit 
force causes unit displacement. By the energy equation (III), 
Art. 53, we infer that unit change in kinetic energy arises when 
unit work is done. The unit of energy is accordingly the same 
as the unit of work. If scientific units are employed, the unit 
of distance is the foot, the unit of force is the poundal, and the 
unit of work and energy is called the foot-poundal. If technical 



WOEK, ENEEGY, IMPULSE 123 

units* are employed, the unit of force is the pound and the unit 
of work and energy is called the foot-pound. Since one pound of 
force = g poundals, we have 

1 foot-pound = g foot-poundals. 

In the French system the unit of distance is the centimeter, the 
unit of force is the dyne, and the unit of work is called the erg. 

Poiver. The question of time does not enter in calculating 
the amount of work done. Power is defined as the rate of doing 
work. The unit in the English system is the horse power, which 
is the equivalent of 550 foot-pounds per second, and in the 
French system is the force de cheval, which is the equivalent of 
75 meter-kilograms per second. 

The relation between the units in the English and French sys- 
tems is exhibited in the following table of equivalents: 

1 foot-poundal = 4.214(10) 5 ergs. 
1 foot-pound = 1.356(10) 7 ergs. 
1 horse power = 1.014 force de cheval. 



PROBLEMS 

1. Compute the energy of a body weighing 300 lb. and moving at the rate 
of 16 ft. per second. Ans , i 2 oo foot-pounds. 

2. A body weighing 10 lb. is thrown upward against gravity. Compute 
the work done upon it by its weight (a) while it rises 10 ft., (6) while it falls 
10 ft - Ans. (a) — 100 foot-pounds ; (b) + 100 foot-pounds. 

If the resistance of the air amounts to a constant force of 2 lb., compute the 
work done by it in both cases. Ans , _ 2 o foot-pounds in each case. 

3. If a body of 10 lb. mass is projected horizontally on a rough plane with 
a velocity of 50 ft. per second, how far will it move before its velocity is reduced 
to 20 ft. per second, the retarding force due to friction being constantly 5 lb.? 

Ans. 65.2 ft. 

* The energy equation (III), Art. 53, was derived under the assumption that scientific 
units are employed, that is, force is equal to the product of mass by acceleration. If 
technical units are used, we have the relation 

Force = mass x acceleration 
g 
and the energy equation becomes 

Work done = change in kinetic energy = — (v 2 — v%). 

2 <7 



124 THEORETICAL MECHANICS 

4. A body weighing 10 lb. falls vertically under gravity against a constant 
force of 1 lb. due to the resistance of the air. How far must it fall in order that 
its velocity may change (a) from zero to 20 ft. per second, (6) from 10 ft. per 
second to 20 ft. per second ? AnSt (a) 6.9 ft.; (6) 5.2 ft. 

5. A mass of 1000 lb. is moving with a velocity of 2 ft. per second, (a) What 
force will stop it in 0.1 second ? (6) What work is done by the force in stopping it ? 

Ans. (a) 20,000 poundals ; (&) 2000 foot-poundals. 

6. Water is to be lifted 150 ft. at the rate of 5 cu. ft. per second. What 
horse power is required ? Ans. 85.2 l-P. 

(1 cu. ft. of water weighs 62.5 lb.). 

7. Compare the power of two men one of whom can do 4000 foot-pounds of 
work per minute and the other (10) 7 ergs per second. 

8. A steam crane lifts 26,280 lb. 150 ft. in 8 min. What is the horse 
Power? AnSt 18.45 hP. 

9. (1) How long will it take a 3 HP engine to raise 12 T. 42 ft. ? (2) From 
what depth will a 22 HP engine raise 13 T. in one hour ? 

10. The monkey of a pile driver weighing 1000 lb. is raised 20 ft. and 
allowed to fall on the head of a pile which is driven into the ground 1 in. by the 
blow. Find the average force exerted on the head of the pile. Ans. 120 tons. 

11. A train of 60 T. runs a mile on a level track at constant speed. If the 
resisting forces are equivalent to the weight of 8 lb. per ton, find the work done by 
the engine. What must be the minimum hP of the engine to attain a speed of 
20 mi. per hour ? Ans% 126 7 foot-tons ; 25.6 l-P. 

12. Suppose the train of the preceding example has steam cut off and brakes 
applied when running 15 mi. per hour. If it stops \ mi. from where the brakes 
were first applied, find (1) the mean resistance ; (2) the time taken to stop the 
train ; (3) the work done by the resisting forces. 

Ans. (1) 687.5 lb.; (2) 2 min.; (3) - 508.2 foot-tons. 

13. A train runs (under the action of gravity) from rest for 1 mi. down a 
plane whose descent is one foot vertically for each 100 ft. of its length ; if the 
resistances be equal to 8 lb. per ton, how far will the train be carried along 
the horizontal level at the foot of the incline ? AnSm \ m i. 1408 yd. 

14. In how many hours would an engine of 18 l-P empty a vertical shaft full 
of water, if the diameter of the shaft be 9 ft., the depth 420 ft., and the mass of 
a cubic foot of water 62.5 lb. ? Ans. 9.8 hr. 

15. The average flow over Niagara Falls is 270,000 cu. ft. per second. The 
height of the fall is 161 ft. What horse power could be developed from the falls 
if all the energy were utilized ? AnSi 4,940,000 nearly. 

16. A particle has been falling for 40 sec. Find (a) the resultant force 
which will stop it in 10 sec; (•&) in 10 ft. Ans . ( fl ) 4 m i b . ; (&) 2 560 m lb. 

17. A particle whose mass is 8 lb., tied to one end of a fine thread, 6 ft. long, 
swings in the arc of a semicircle. Find its kinetic energy and velocity as it passes 
through the lowest point. Ans% 48 f 00 t-pounds ; 8 V6 ft. per second. 



WORK, ENERGY, IMPULSE 125 

18. The piston of a steam engine is 15 in. in diameter, its stroke is 2£ ft. 
long, and it makes 40 strokes per minute. If the mean pressure of the steam is 15 
lb. per square inch, what work is done by the steam per minute and what is the 
horse power of the engine ? ^ns. 265,072.5 foot-pounds ; 8.03 HP. 

19. What must be the length of the stroke of the piston of an engine, the 
surface of which is 1500 sq. in., which makes 20 strokes per minute, so that with a 
mean pressure of 12 lb. on each square inch of the piston, the engine may be of 
80 horse power ? AnSt 71 f t# 

20. A hammer weighing a lb. , and moving with a velocity v, strikes a nail. 

How far will the nail be driven if it offers a resistance r? Q av 2 

Ans. in. 

rg 

21. The diameter of a piston head is 1 ft., the steam pressure 20 lb. per square 
inch, and the length of the stroke 3 ft. How many strokes per minute must the 
engine make to raise 2 cu. ft. of water per second from a depth of 400 ft., assuming 
that 0.02 of the energy is lost by friction ? (1 cu. ft. of water weighs 62.5 lb.) 

22. A man who weighs 140 lb. walks up a mountain path at a slope of 30° 
to the horizon at a rate of 1 mi. per hour. Find his rate of working in raising his 
own weight in horse power, 

23. An automobile, weighing 1 T., can run up a hill of 1 in 60 at 8 mi. an 
hour. Taking the resistance due to friction as -^ of the weight of the car, find at 
what rate it could run down the same hill, assuming the horse power developed by 
the engine to remain the same. 

24. Assuming that a man in walking raises his center of gravity through a 
vertical height of 1 in. at every step, find at what horse power a man is working in 
walking at 4 mi. an hour, his stride being 33 in., and his weight 168 lb. 

56. Impulse. A second fundamental equation for rectilinear 
motion is obtained by integration of the force equation, 

rr dv 
F = m—- 

dt 

Multiplying both sides by dt, interchanging members, and inte- 
grating between the limits t and t' for the time, we obtain 



J I mdv = 1 
to Jk 



Fdt. 



The second member is called the time-integral of the force 
F. If v and v' are the values of the velocity for t = t Q and 
t = t f , respectively, then we may write 

(1) mv' —mv Q = f Fdt, 

or, introducing momentum, 

(VI) Change in momentum = time-integral of force. 



126 THEORETICAL MECHANICS 

For a reason now to be explained, the equation (1) is called the 
impulse equation. Changing the limits in (1) from t == to t—t, 
it becomes 



(2) mv — mv = j Fdt. 



In this equation the force F is assumed continuous. As a con- 
sequence velocity will necessarily change continuously with the 
time, and therefore a force cannot cause a sudden change of 
momentum. The phenomenon of sudden changes in velocity, 
such as are produced by blows, is, however, frequently observed. 
Such changes are said to be caused by impulses or impulsive forces. 
That is, impulses cause changes of velocity in a time too short to 
be measurable. In this phenomenon the change of momentum, 
m V— mv, where v and V are respectively the initial and final 
velocities, may be observed. For this reason an impulse is said to 
be measured by the change of momentum it causes, and in scien- 
tific units is set equal to this. That is, using R for impulse, 

(3) R=m V— mv. 

This equation may be regarded as a limiting case of the impulse 
equation, and the latter designation is derived from this fact. 
For if F is the mean value of the force F in the time t, we have 

J' Fdt = Ft (Calculus, p. 358). The impulse equation (2) may 
o 

now be written mv _ mv = Ft. 

To apply this to a sudden change in velocity, we may let t 
diminish and F increase in such a way that the product Ft re- 
mains finite and approaches a limiting value, namely, 

limit v, Tr 

r,Jit = mV— mv, 

F'and v being as before the final and initial velocities. Compar- 
ing with the definition (3), we have 

„ limit ™ 
R = t;=0 J<t, 

that is, from the present point of view, a sudden change in 
momentum may be roughly regarded as caused by a very great 
force acting for a very small time, and the corresponding impulse 
is measured by their product. 



WORK, ENERGY, IMPULSE 127 

Dimensions of impulse and momentum. Momentum has been 
defined (Art. 43) as the product of mass by velocity. Its dimen- 
sions in terms of the fundamental units of mass, length, and time 
are expressed by the equation 

tv/t , mass x length 
Momentum = : s 

time 

From the preceding definition of impulse it is clear that its dimen- 
sions are the same as those of momentum. 

57. Impact. Problems in impact or collision of solids furnish 
examples of impulses. 

Consider, for example, the impact of a solid elastic sphere 
upon an elastic plane surface, the direction of motion being along 
the normal to the plane. The phenomena during 
contact may be described as follows : 

1. The sphere is compressed until its velocity 
is zero. 

2. The sphere then assumes its original shape 



and a certain final velocity. 

Obviously, the change in momentum in each 
of the two stages described may be regarded as caused by an 
impulse, and we shall have by definition, if m is the mass of the 
sphere, 

R = impulse of compression = mv, 

if v is the original velocity, and 

R' = impulse of restitution = mv\ 

if v f is the final velocity. 

Now it is a fact observed by Newton, that the final and initial 
velocities are in a ratio which depends upon the substances in 
contact, and not upon the velocity of impact. This constant 
ratio, called the coefficient of elasticity, we denote by e, and hence 
v' — — ev, the negative sign indicating reversal in direction. Con- 
sequently, the impulses satisfy the relation 

(4) R< = - eR. 

The coefficient e is less than unity, if the solid and plane sur- 
faces are not perfectly elastic. There is accordingly a loss of energy 




128 THEORETICAL MECHANICS 

by the impact. No exception, however, is here afforded to the 
doctrine of the conservation of energy. The accompanying phe- 
nomena of heat, light, etc., indicate a transformation of energy 
corresponding to the lost energy of motion. 

Oblique impact. If the direction of approach be 

inclined at an angle to the normal, and if the 

sphere and plane are smooth, then no change of 

N motion occurs along the surface of the plane, and 

the preceding discussion applies without change to 

• the normal components of the velocities. That is, 

(5) v n ' = - ev n , R' = mvj, R = mv n , R' = - eR. 

Since the tangential components (along the plane surface) are 
equal, that is, v/ = v t , we shall have 

(6) tan & = ^4 = — = -tan 0. 

If the solids are imperfectly elastic, tan 6 r > tan 0, hence 6' > 6, 
and the path is bent away from the normal. 

Illustrative Example. A bullet weighing half an ounce is fired with a 
speed of 2000 ft. per second from a rifle weighing 10 lb. If the rifle kicks back 
through 3 in., find the average pressure applied by the shoulder in bringing it 
to rest. 

Solution. Since the impulse acting upon the gun is equal and opposite to the 
impulse acting upon the bullet, we have the relation 



where »i, v and m', v' denote the masses and velocities of the gun and bullet, 
respectively. Substituting the values of m, m,' v', we find 

v = z -£ ft. per second. 

If F denotes the average pressure exerted by the shoulder, we have from the work 
equation, 

f- 1 = *■ io •(¥)»; 

whence 

F = lip poundals = 24.42 lb. 

PROBLEMS 

1. An arrow shot from a bow starts off with a velocity of 120 ft. per second. 
With what velocity will an arrow twice as heavy leave the bow if sent off with 
three times the force ? Ans. 180 ft. per second. 

2. A ball falls from rest at a height of 20 ft. above a fixed horizontal plane. 
Find the height to which it will rebound, e being |, and g being 32. Ans. \1\ ft. 



WORK, ENERGY, IMPULSE 129 

3. A ball falls from a height h on to a horizontal plane, and then rebounds. 
Find the height to which it rises in its ascent. Ans. e 2 h. 

4. A ball is projected from the middle point of one side of a billiard table, so 
as to strike first an adjacent side, and then the middle point of the side opposite to 
that from which it started. Find where the ball must hit the adjacent side, its 

length being b. ^ ns> At tne Stance from the end nearest the opposite side. 

1 + e 

5. A rifle weighing 3 lb. is discharged while lying on a smooth horizontal 
table. The weight of the bullet is 2 oz., and it leaves the gun with a velocity of 
1400 ft. per second. What is the impulse of the kick ? 

6. A man weighing 180 lb. jumps from a boat weighing 110 lb. into a boat 
weighing 160 lb. If the boats are initially at rest compare their velocities after 
the jump. 

7. A ball whose mass is 5J; oz. is moving at the rate of 100 ft. per second 
when it is struck by a bat in such a way that immediately after the blow it has a 
velocity of 150 ft. per second in a direction making an angle of 30° upward from the 
horizontal. Assuming the horizontal velocity to be reversed by the blow, find the 
value and direction of the impulse. 

8. A shot of 700 lb. is fired with a velocity of 1600 ft. per second from a 
35-T. gun. Find the velocity with which the gun recoils, neglecting the weight of 
the powder. If the recoil is resisted by a steady pressure equal to the weight of 
10 T., through what space will the gun move ? Ans. 14 } ft. per second ; 11/g ft. 

9. A particle falls from a height h upon a fixed horizontal plane. If e be the 
coefficient of restitution, show that the whole distance described by the particle 

before it has finished rebounding is h e , and that the time that elapses is 



A 



2 h 1 + e 
g 1- e 



10. A smooth elastic ball is projected horizontally from the top of a tower 
100 ft. high with a velocity of 100 ft. per second, and after one rebound describes a 
horizontal range of 40 ft. Find the coefficient of elasticity. Ans. £ z . 

11. Two equal scale pans, each of mass M, are connected by a string which 
passes over a smooth peg, and are at rest. A particle of mass m is dropped on one 

of them from a height — , the coefficient of elasticity being e. Find the velocity of 

the scale pan after the first impact. Ans. — — — (1 + e). 

2 M + m 

12. Show that a billiard ball of any elasticity, struck from any point on the 
table, and returning to the same point after impinging against the four sides in 
order, describes a parallelogram, with sides parallel to the diagonals of the table. 

13. A heavy elastic ball drops from the ceiling of a room and after twice 
rebounding from the floor reaches a height equal to one half that of the room. 
Show that its coefficient of restitution is y/\. 



130 THEORETICAL MECHANICS 

14. A heavy elastic ball falls from a height of n ft. and meets a plane inclined 
at an angle of 60° to the horizon. Find the distance between the first two points 
at which it strikes the plane. Ans. 2 V3 ne(l + e). 

15. The sides of a rectangular billiard table are of lengths a and b. If a ball 
of elasticity e be projected from a point in one of the sides, of length b, to strike 
all four sides in succession and continually retrace its path, show that the angle of 
projection with the side is given by ae cot 6 = c + ec', where c and d are the parts 
into which the side is divided at the point of projection. 

16. A smooth circular table is surrounded by a smooth rim whose interior 
surface is vertical. Show that a ball of elasticity e projected along the table from a 



point in the rim in a direction making an angle arc tan \ with the radius 

v 1 + e + e 2 

through the point will return to the point of projection after two impacts on the 

rim. Prove also that when the ball returns to the point of projection its velocity is 

to the original velocity as e? : 1. 

58. Force-moments in a plane. The discussion of this article 
will enable us to work out a second* integration of the force 
equations. 

Multiply the first of these equations by y, the second by #, and 
subtracting, we get 

The second member, xF y — yF x , has a simple geometrical sig- 
nificance. If the directed line PQ represents the force whose 
axial components are (JP^, Fy) and whose point of application is 
P(x, y), then the coordinates of the point Q are at once seen from 
figure a to be (x + F x , y -f- Fy), for 

ON= OM+MN=x + F x , NQ = MP + SQ = y + F y . 

* The energy equation, Art. 53, is to be regarded as a first integration of the force 
equations. For writing these in the form 

mv x ^ = F mv ^k = f v , 
dx x y dy r 

multiplying the first by dx, the second by dy, and adding, gives 
m(v x dv x + v y dv y ) = F x dx + F y dy. 
Integrating, we obtain 

i m(Vx 2 + Vy2) + C = J (F x dx + F y dy) . 
But v 2 — v x 2 + v y 2 , and hence the result is a form of the equation in question. 



WORK, ENERGY, IMPULSE 



131 



In figure b the area of the triangle OPQ is (Analytic Geome- 
try, p. 42) 



x r »J 



Q 

S 



% M 



(a) 



N 




w 



(3) \\x(y + FJ - y(x + J^] = JO**;- ^). 

But the area of the triangle also equals \pF, if p is the altitude 
drawn from the origin upon PQ. 

(4) r.xF y -yF x =pF. 

The product pF is called the moment of the force F with 
respect to the origin. The point is called the center of moments. 
The perpendicular distance p is called the lever arm. Hence 

Force x lever arm = force-moment. 

Equation (4) gives the expression for force-moment in ana- 
lytic form. That is, 

(VII) Force-moment with respect to the origin = xF y — yF x , 

where the axial components of the force are (i^, Fy) and the 
point of application is (#, y). 

Sign of force-moment. The area of OPQ is positive when the 
order 0, P, Q of the vertices on the perimeter is counter-clock- 
wise ; negative, if clockwise. 
Hence force-moment is positive 
if the force acts to cause posi- 
tive rotation (counter-clockwise) 
about the point ; otherwise, 
negative. 

Evidently, if p = 0, the force- 
moment vanishes, that is, if the 
line of action of the force passes 
through the center of moments the force-moment vanishes. Force- 
moment is unchanged if the point of application is displaced along 




Positive 



Negative 



132 



THEOBETICAL MECHANICS 




the line of action, the magnitude and direction of the force re- 
maining the same. For F and p in (4) both remain constant if 
the point of application is so displaced. 

Consider next two concurrent forces 
F(F X , F y ) and F'(FJ, F y f ), the common point 
of application being (x, y}. Then, by (VII), 

Moment of F= xF y - yF x , 

Moment of F = xF y ' - yFJ '. 
Adding, 

(5) Moment of F + Moment of F' = x(F y + F y ! ) - y(F x + FJ). 

Let R be the vector resultant of i^and^. Then the axial 
components of R are (F X + FJ, F y + F y '~). The second member 
of (5) is therefore the moment of R. Hence 

(6) Moment of F -\- moment of F — moment of R. 

That is, the sum of the force-moments of two concurrent forces is 
equal to the moment of their resultant. This principle is general; 
it can be extended to any number of concurrent forces and leads 
to the important 

Theorem of Moments. The algebraic sum of the force- 
moments of any number of concurrent forces with respect to any 
center equals the force-moment of their resultant. 

59. Moment of momentum. Consider the first member of (2), 
Art. 58. This expression is a time-rate. For it is easily seen that 

,., ^ / dv v dv r \ d , >, 

(1) mix — u — y — - ) = ■ — (x-mv v —y'mv r ). 
y J \ dt u dtj dV y U J 

Hence the equation (2), Art. 58, takes 

the form 

d 



( 2 ) J t (~ x ' mv y 




y • mv x ) = xF y - yF x . 

This result is called the moment equation 
for a reason now to be explained. 

The expressions mv x , mv y in the first member of (2) are the 
axial components of the vector momentum mv. The expression 

(3) x • mv y — y • mv x , 

being entirely analogous to 



x-F„ 



y 



F„ 



WOKK, ENERGY, IMPULSE 133 

is called the moment of momentum with respect to the origin. The 
moment of momentum is of course variable in any general motion, 
and is a function of the time. Equation (2) therefore gives the 
relation, 

(VIII) 
Time-rate of change of moment of momentum = force-moment 

In the equations of motion, F x and F y are the axial com- 
ponents of the complete resultant of all forces acting. Hence in 
(VIII) the force-moment is the resultant force-moment (Theo- 
rem of Moments). 

60. Angular momentum. The expression for moment of 
momentum is simple if polar coordinates are used. In the figure, 
let p be the perpendicular distance from the center of moments 
to the tangent to the path. Then p is the lever 
arm of the vector momentum. Hence, by Art. 
59, 



\ 



(1) Moment of momentum = mv -p. 



de 



But in the figure, p = p sin yfr = p ■ p— . y4 

CIS 




V 



Also, since v = — , (1) becomes 

(2) Moment of momentum = m—- p 2 - — = mp 2 — . — = mo 2 — . 
V J J dt r ds H dt ds ^ dt 

Now (see Art. 22), mp 1 = moment of inertia of the particle 

d6 * 

with respect to O(=_Z ), and since —= angular velocity = w, 

we obtain the result, 

(3) Moment of momentum = I Q (o. 

If this result is compared with the definition of momentum 
(mv), it is seen that moment of inertia corresponds to mass, and 
angular velocity to linear velocity. From the introduction of 
angular velocity, moment of momentum is often called angular 
momentum, and momentum proper, linear momentum. We thus 
have the definitions 

Linear momentum — mass x linear velocity, 
Angular momentum = moment of inertia x angular velocity. 




134 THEORETICAL MECHANICS 

Moreover, the results of Arts. 44 and 59 give the equations, 
(4) Force = — (linear momentum) ; 

Out 

Force-moment = — (angular momentum). 

Integration of the last equation will give the result: 

Change in angular momentum = time-integral of force-moment. 

This result is to be regarded as a second integral of the force 
equation, the other being the energy equation already found. 
Comparison should be made with (VI) for 
rectilinear motion. In the latter, momentum 
has been replaced by angular momentum and 
force by force-moment. 

Illustrative Example. As an application, consider 
central motion, that is, motion caused by a force con- 
stantly directed towards a fixed center 0. 

Solution. The force-moment with respect to O is zero. Hence, choosing 
for origin, (4) gives 

fjf) dft 

mp 2 — = constant, or p 2 — = constant. 

H at H dt 

If u is the area of any sector A OP, then 

u = 1 ( P 2 dd, and — = \p 2 - = constant. 
J dt dt 

The derivative — , that is, the time-rate of change of the sectorial area AOP, is 
dt 

called the areal velocity. Hence in any central motion the areal velocity is constant ; 

or also the radius vector siveeps over sectors of equal area in equal times. 

61. Fundamental equations. The preceding sections have led 
to three types of equations, namely: 

Rectilinear Motion. Curvilinear Motion. 

Force Equation, Force Equations, 

Energy Equation, Energy Equation, 

Impulse Equation. Moment Equation. 

Problems in motion depend for their solution largely upon 
these equations, and their application has been seen to be of 
fundamental importance. The moment equation will be used in 
a later chapter. 



WORK, ENERGY, IMPULSE 



135 



62. Formulas in dynamics of a particle. For convenience 
of reference the chief results in connection with the kinetics of a 
material particle are collected here. In every case F denotes the 
resultant of all forces acting on the particle. 

F T = sum of rE-components of all forces. 

F y = sum of y- components of all forces. 

F z = sum of 2-components of all forces. 

F t = sum of tangential components. 

F n = sum of normal components. 

F p = sum of components along radius vector. 

F = sum of components perpendicular to radius vector. 

Force Equations : 

d 2 x dv r dv r -n 

m— = m— -z = mv x —^ = F x , 
at 2, at ax 



dv v dv v 77 

dt 2 dt y dy 



dfy 
dt 2 
d 2 z 



dv z dv s 

m—- =m — ? = mv z — — 
dt z dt dz 



F. 



dv dv xr 

m— =mv—- = F t , 
dt ds 



mv 



B= F " 



4S-<f)'H 



p 1 



-TAP 



16\ 



pdtX dtj 



= F a 



^Rectangular 
Coordinates. 



Intrinsic Equations. 



Polar 
Coordinates. 



Work Integrals : 

Work = / (F x dx + F y dy\ Rectangular Coordinates. 

Work = / F ( ds, Intrinsic Equations. 

Work = iCF p dp + F e pdd), Polar Coordinates. 

Energy Equation : 

Work done by all forces acting = \ mv 2 — j- mv 2 , 
initial velocity = v , final velocity = v. 



136 THEOEETICAL MECHANICS 

Impulse Equation : 
Impulse = mV — mv, 

initial velocity = v, final velocity = V. 

Moments : 

Force-moment = xF y — yF x , 

point of application = (x, y), center of moments = (0, 0). 



Moment Equation : 
e-moment = — 
I = moment of inertia and co = angular velocity with 



Force-moment =— (angular momentum) = — (i^®), 

OlV Olv 



respect to (0, 0). 



CHAPTER VI 

MOTION OF A PARTICLE IN A CONSTANT FIELD 

63. Field of force. A region in which force is known to exist 
is called a field of force. A body free to move when placed in 
such a field will in general not remain at rest. The acceleration 
which will be imparted to a material particle at any point in a 
field of force must be regarded as characteristic for the field at 
that point. Upon particles of different masses m v ra 2 , etc., at the 
same point the field will exert different forces. These forces will 
have the same direction, but different magnitudes, namely, m x f 
m 2 f etc., if/ is the acceleration at the point. Consequently, if we 
know the acceleration due to a field at each of its points, we know 
the force the field will exert upon any material particle. For this 
reason a field of force is said to be determined ivhen the vector accel- 
eration at every one of its points is known. 

In general the acceleration due to a field of force is different at 
different points of the field. When the acceleration is the same at 
every point, the field is said to be constant or uniform. A familiar 
and important example of a constant field is afforded by consider- 
ing the earth's attraction, that is, the force of gravity, in any small 
region. All bodies are attracted towards the center of the earth. 
Particles falling freely from rest will describe rectilinear paths 
which may be regarded as parallel if the region under considera- 
tion is small, and experiment shows that the acceleration at every 
point of such a region is constant. 

64. Rectilinear motion under a constant force. This type of 
motion is very important. We consider the motion of a material 

-d 1 ~ —* 

particle along a straight line, the force causing the motion being 
directed along that line. 

Assume a positive direction and an origin of distance on the 
path, and let s denote the distance from the origin to any position 

137 



138 THEORETICAL MECHANICS 

P. The driving force F is then mf or — mf according as its 
direction is positive or negative. From the force equation we have 

as , n as /» 

™- = ±m/, or - = ±/. 

Integrating, using s and v to determine the initial position 
and velocity, we derive the result: 

The velocity and position of a material particle moving along a 
straight line under a constant force are given by 

(I) V = V ±ft, S = S + V t±lft 2 , 

in which f denotes the constant acceleration, and v and s Q determine 
the initial velocity and position, respectively. The positive or nega- 
tive sign holds according as the distance and force have the same or 
opposite directions. 

The energy equation is 

l mv i _ i mv 2 — j ]?$$ _ ]?q s _ § ^ __ ± m f( s _ g() ). 

Hence if d = s — s = distance moved, this equation gives, by solv- 
ing, 

(II) v^ = v ^±2fd, 

expressing the final velocity in terms of the initial velocity and 
distance moved. 

When the initial velocity is opposed to the force, from (II), 

v 2 = v 2 — 2fd, and (I), v = v —ft. Hence v = when t — -A and 

v 2 £ 

then d = -&-, the velocity constantly diminishing until this position 

is reached, and thereafter constantly increasing. 

Analysis of (I) is important. If v = s = 0, then 

v = ±ft, s=±±M ...f=±M, 

formulas giving the velocity and distance due to the force only. 
To obtain (I) and (II), we add on the initial velocity to get the final 
velocity, and the distance ( = v fy due to it and the initial distance 
to get the total distance. 



MOTION OF A PARTICLE IN A CONSTANT FIELD 139 



ILLUSTRATIVE EXAMPLES 

1. Bectilinear motion under gravitation. In this case f = g. We distinguish 
two problems. 

Particle projected vertically upwards. Taking the upward direction as 
positive, and the origin as the point of projection, we have in (I), s = 0, and 
obtain 

(1) v = v - gt, s = v t - \ gt 2 , 

in which v is the velocity of projection. The highest point A is reached 

when v = 0. Then t = — , and s = OA = ^-. This is therefore the 
greatest height. ^ ^ 

Particle falling freely. Choosing the downward direction as positive, we 
obtain from (I), ■" 

(2) v = v + gt, s = s + v t + \ gt 2 . 
If the particle falls from rest at the origin, then v = 0, s = 0, and we get 

(3) v = gt, s = \gt 2 , v 2 = 2gs ; 
Hence the velocity acquired in falling freely from rest a distance h q 

equals V2 gh. 

2. Atwood's machine. Let the figure represent two masses mi and ra 2 sus- 
pended by an inextensible thread passing over a smooth pulley. 
The motion of the system is known if the motion of one of the 
particles m 2 is known, that is, if m 2 descends with an accelera- 
tion /. In other words, if the acceleration of m 2 is /, the ac- 
celeration of mi is — /• Denoting by T the tension in the 
thread, that is, the pull of the particle on the thread, the re- 
sultant force acting on m 2 is m 2 p r — T, and on mi is mig — T. 

Hence the force equation gives 

J wi2<7 — T = m 2 f, 
[niig - T = - mi/. 

Eliminating T and solving for /, we find 

-_ mi — mi 

m 2 + mi 

The acceleration is constant and the equation of motion is found by substitution 

in (I). 

Eliminating / from the force equations, we obtain the tension in the thread, 

namely, _ 

ji _ 2mim 2 

mi + w 2 

PROBLEMS 

1. How long will it take a body to fall from rest through 625 ft.? Find the 
velocity acquired. How far does it fall in the fourth second ? (g= 32.) 

Ans. 2£ sec; 200 ft. per second; 112 ft. 




140 THEOKETICAL MECHANICS 

2. How long will it take a body to acquire a velocity of 260 ft. per second 
falling from rest ? An ^ jy> sec> 

3. (1) How high will a stone rise which is projected upwards with a velocity 
of 256 ft. per second ? (2) What is its position, direction of motion, and velocity 
at the end of the tenth second ? (g = 32.) 

Ans. (1) 1024 ft. (2) 960 ft. high ; v = 64 ft. per second downwards. 

4. Compare the momentum of a 3-lb. weight after falling 30 ft. with that of a 
half-ounce bullet having a velocity of 2000 ft. per second. Ans. 24 V30 : i- 2 ^. 

5. With what velocity must a body be projected downwards that it may over- 
take in 10 sec. another which has already fallen through 100 ft. ? 

Ans. 90 ft. per second. 

6. A body is projected upwards with a velocity of 80 ft. per second. How 
long will it be in returning to the starting place ? With what velocity will it 
return ? (g = 32.) AnSm 5 sec> . 80 f t . per second. 

7. A particle has an initial velocity of 125 cm. per second, and an acceleration 
(1) of 10 cm. per second each second ; (2) of— 10 cm. per second each second. 
How long will it take in each case to move over 420 cm. ? Explain the results. 

Ans. (1) 3 sec. or — 28 sec. ; (2) 4 sec. or 21 sec. 

8. The velocity of a particle moving in a constant field is a cm. per second 
at the end of c seconds, and b cm. per second at the end of (c + 1) sec. What was 
the initial velocity and acceleration ? Ans , Vq== a + (a-b)c,f= b - a. 

9. The sum of the two weights of an Atwood's machine is 12 lb. The heavier 
weight descends through 128.8 ft. in 8 sec. What are the values of each weight? 
{g = 32.2.) AnSm 6.75 lb., 5.25 lb. 

10. A 2-lb. weight carried on a spring balance in a balloon has an apparent 
weight of 2.4 lb. when the balloon is ascending. What is the acceleration of the 
balloon ? What should the body weigh if the balloon is descending with an accel- 
eration of 10 ft. per second ? AnSm 33.4 ft. per sec. ; -^ lb. 

11. A mass of 12 lb. rests on a smooth horizontal table. A second mass of 
1.5 lb. is attached to the first by means of a cord passing over the edge of the table. 

Find the following : 

(1) The acceleration of the system. Ans. - 3 ^- ft. per second. 

(2) The space described in 3 sec. 16 ft. 

(3) The velocity attained at the end of 5 sec. x f a ft. per second. 

(4) The force on the string. L |- 8 poundals. 

(5) The time required for the system to move 120 ft. 3\Af sec. 

12. Two unequal masses are connected by a weightless inextensible string 
passing over a smooth peg. What must be the ratio of the masses that the system 
may move through 24 ft. in 3 sec. from rest ? Ans. 5 : 7. 

13. A train passes another on a parallel track, the former having a velocity of 
45 mi. an hour and an acceleration of 1 ft. per second per second, the latter a 
velocity of 30 mi. an hour and an acceleration of 2 ft. per second per second. How 
soon will the second be abreast of the first again, and how far will the trains have 
moved in the meantime ? 



MOTION OF A PAKTICLE IN A CONSTANT FIELD 141 

14. A body is dropped from a balloon at a height of 70 ft. from the ground. 
Find its velocity on reaching the ground, if the balloon is (a) rising, (6) falling 
with a velocity of 30 ft. per second. 

15. A stone is dropped into a well, and the sound of the splash reaches the 
top after 9 sec. Find the depth of the well, the velocity of sound being 1100 ft. 
per second. 

16. A body whose mass is 5 lb. , moving with a speed of 160 ft. per second, 
suddenly encounters a constant resistance equal to the weight of | lb., which 
lasts until the speed is reduced to 96 ft. per second. For what time and through 
what distance has the resistance acted ? 

17. A body falls freely from the top of a tower and during the last second of 
its flight falls |f of the whole distance. Find the height of the tower. 

Ans. 100 ft. 

18. Two scale pans of mass 3 lb. each are connected by a string passing over 
a smooth pulley. Show how to divide a mass of 12 lb. between the two scale pans 
so that the heavier may descend through a distance of 50 ft. in the first 5 sec. 

Ans. In the ratio 19 : 13. 

19. A string hung over a pulley has at one end a mass of 10 lb. and at the 
other end masses of 8 and 4 lb., respectively. After being in motion for 5 sec. the 
44b. mass is taken off. Find how much farther the masses go before they come 
t0 rest - Ans. 29 ft. 9 in. nearly. 

20. If the string in an Atwood's machine can bear a strain of only \ the sum 

g 
of the two weights, show that the least possible acceleration is —= . Find the least 

ratio of the larger to the smaller weight. 

21. A mass m pulls a mass m' up an inclined plane, inclination a, by means 
of a string passing over a pulley at the top of the plane. Show that the accelera- 



m + m' 

22. A mass of 6 oz. slides down a smooth inclined plane, whose height is half 
its length, and draws another mass from rest over a distance of 3 ft. in 5 sec. along 
a horizontal table which is level with the top of the plane over which the string 
passes. Find the mass on the table. Ans. 24 lb. 10 oz. 

23. A weight P is drawn up a smooth plane inclined at an angle of 30° to the 
horizon, by means of a weight Q which descends vertically, the weights being con- 
nected by a string passing over a small pulley at the top of the plane. Find the ratio 
of Q to P if the acceleration is ^. Ans. Q = P. 

24. A juggler keeps three balls going with one hand, so that at any instant two 
are in the air and one in his hand. Find the time during which a ball stays in his 
hand if each ball rises to a height of a ft. 

25. A stone dropped into a well is heard to strike the bottom in t sec. Find 
the depth of the well, the velocity of sound being a ft. per second. 



Ans. 



■J<u+f—S= 

\ 2 fj V2 ij 



142 THEOEETICAL MECHANICS 

26. The two masses of an Atwood's machine are 8 and 10 lb., respectively, and 
the string is clamped so that no motion can take place. If the string is sud- 
denly undamped, find the change in pressure exerted on the pulley. 

27. A mass of 10 lb. resting on a smooth inclined plane, inclination 30°, is 
connected by a string passing over a pulley at the top of the plane to a mass of 
10 lb. hanging vertically. Find the tension in the thread (1) when the weight 
on the plane is held fixed, (2) when the hanging weight rests on a support, (3) when 
both weights are free to move. 

65. Curvilinear free motion. We first prove the 

Theorem. If a free material particle is projected into a constant 
field in a direction oblique to the direction of the force of the field, 
the path will be a parabola. 

Let the acceleration due to the field be / and its direction 
opposite to the direction of the P-axis. The axial components of 
the force at any point are 

F x = 0, F y = - mf. 

Hence the rectangular force equations are 

(i) w <g = o, »5=-»/. 

Let the initial position be (# , «/ ), the initial velocity v , and 

the angle which v makes with 
r| the X-axis a. Then the com- 

ponents of the initial velocity 
are v cos a and v sin a, re- 
spectively. Integrating 
equations (1) and determining 
— ^ the constants by the given 
initial conditions, we obtain 

(2) x = x 4- v cos a • t, y = y + v sin a-t— }ft 2 . 

Eliminating £, the equation of the path is 

( 3 ) y - jf = tan '« O - x o) - g - 2 { os2 - O- %) 2 > 

which is the equation of a parabola with its axis parallel to the 
!F-axis. Q. e. d. 

The distance moved through in the direction of the force is 
— Qy — «/ ). Hence the energy equation gives 

W ^ = V_2/(y-y ). 




(%o,yo) v cosa 



MOTION OF A PARTICLE IN A CONSTANT FIELD 143 



The applications of the preceding theorem are mainly to prob- 
lems of the motion of projectiles near the surface of the earth. 
Neglecting the resistance of the atmosphere and the variation of 
the force of gravity, the circumstances of the motion are given by 
(2) where / ' = g. If the origin of coordinates is the initial posi- 
tion, the X-axis horizontal, and the positive direction of the 
J^-axis upwards, the equations of motion are 

(III) a? = v cos at, y = v sin a • t - ^ gP. 

and the equation of the path and the energy equation become, 

respectively, y = tanaa ,_ 9 &, 

2 vq 2 cos 2 a 

(IV) v 2 = vo 2 - 2 gy. 

The projectile reaches its greatest height J? when the velocity 
in the direction of the Z~-axis is zero. Then v y = 0, and 
v = v x = v cos a. Making these 
substitutions in (IV), and setting 
y = H, we obtain 

2gH=v* 



>2„ — 



— vS cos* a = Vrf sur a 



sin* a 




(5) ... JT=J^ 

The time of flight T is the time elapsed when the projectile 
again reaches the X-axis. Setting y = in (III), we obtain 

= v sin a • t — -gt 2 , whence t = (at 0), 



or t 



(6) 



2 v sin a 
9 

9 



The horizontal range R is the intercept OA on the X-axis, the 
value of x when t= T. Setting t = T in (III), we obtain 



v cos a ' T 



z v * sin a cos a 



_ v 2 sin 2 a 
9 



co 



9 



From (7) it is obvious that the maximum range for a given 
velocity of projection results when sin 2 a = 1 or a = 45°. 



144 



THEORETICAL MECHANICS 



have 



ILLUSTRATIVE EXAMPLES 

1. Find the range on an inclined plane through the origin with an inclination 0. 
Solution. The equation of 00 is y = tan • x. Substituting from (III), we 

v sin « • t gt 2 = tan vo cos a • t, 



or solving, 



t = 



2 vo (sin a — tan cos a) 2-u sin (a — 0) 



which gives the time when the projectile will strike the plane at C. 

Since 00= OM sec 0, and OM is 
the value of x(= Vocos « • t) when £ has 
the value just found, we readily de- 
duce the result, 

2 vo 2 cos a • sin (a - 0) 

COS 2 

The velocity at C is" (by IV) 

^ v 2 = v 2 -2g.MO =v 2 -2g-OC sin 0. 

The angle of impact with the plane, namely, y = r — 0, is readily found. For 
tan t is the slope of the parabola. Substituting the value of OM (= x) already 
found, in this, and reducing, gives 

sin (0 - a) 




tan 7 = 



cos (0 + a) +2 sin tan cos a 



2. Required the elevation in order that the projectile may pass through a given 
point, the velocity of projection being a given constant. 

Solution. Let the given point be Q(xi,yi). Since this point lies on the 
parabola (IV), we have n 



ijx = tan a x\ 



Xy 



tan 2 a + 1 



, substituting and col- 



2 vo 2 cos 2 a 

from which a must be determined. Since cos 2 a 
lecting gives the equation 

(1) gx x 2 tan 2 a - 2 v 2 x x tan a + (2 v^yx 4- gx x 2 ) = 0, 

in which tan a is the unknown. 

Since the equation (1) has in general two roots, the point Q may be reached in 




AX 



two ways. To cover all cases, find the discriminant of (1). Since A = gxp, 
B = — 2v %i, C = 2v 2 y-L + gxi 2 , we have for the discriminant 

A = B 2 - 4 AG = 4 V x x 2 - 8 vfgxfyx - 4 g 2 x^ = 4 x x 2 (v^ - 2 v Q 2 gyi - g 2 X! 2 ). 
Hence A = if v t _ 2 ^2^ _ ^2 = , 



MOTION OF A PARTICLE IX A CONSTANT FIELD 145 

or, omitting the subscripts, 

(2) * 2 =^_1^, 

9 1 9 

v 2 
The locus of this equation is the parabola A'BA. where OA = 2 • OB =-£-. This 

9 
parabola is called the bounding parabola. The final statement obviously is as 

follows : 

If Q is within the bounding parabola, two parabolic paths pass through it 
(A >0 and (1) has real and distinc.t roots). 

If Q is on the bounding parabola, one parabolic path passes through it, and this 
will touch the bounding parabola at Q (A = and the roots of (1) are real and 
equal) . 

If Q is without the bounding parabola, no trajectory passes through it. 

Applied to gunnery, the interpretation of the results is as follows : The 
region covered by a projectile with a given muzzle velocity is the interior of a parabo- 
loid of revolution whose axis is vertical (obtained by revolution of A'BA around 
OY). Any point within the paraboloid may be hit in two ways. 

In terms of the greatest vertical height h, which can be attained with the 
given initial velocity vo* the equation of the bounding parabola takes a simple form. 
The height h is attained when the particle is projected vertically upwards, and is 

Substituting this value of v 2 in (2), the equation of the bounding parabola becomes 

(3) 4:hy + x* = 4h 2 . 

Another convenient form of this equation is obtained by introducing the greatest 
horizontal range r. This is found from (7), Art. 65, by putting sin 2« = 1, whence 

(4) . r = l ^ = 2h, 

9 

and the equation of the bounding parabola may be written 

(5) 2,=A(r2-*2). 

r 2 

3. A man can throw a ball 100 yd. on a horizontal plane, (a) Find the 
highest point that he can hit on a vertical wall 35 yd. away. (6) If he stands on a 
cliff 150 feet high, how far from the base can he throw ? 

Solution. Evidently in either case the greatest distance will be the point 
where the bounding parabola cuts the given plane. We have r= 100 (yd.), and 
hence the equation of the bounding parabola is 

(1) 200 y = 10,000 -z 2 . 

(a) This parabola will cut the vertical line x — 35 at the height 

10000 - 1225 = 4388d 

y 200 J ' 

which is the highest point on the wall that he can hit. 

(V) Since the horizontal plane is 50 yd. below the top of the cliff, we substitute 
in (1) y = - 50, and find a: = 141.4 yd.. 

which is the greatest distance from the base that he can throw. 



146 THEOEETICAL MECHANICS 



PROBLEMS 

1. A gun is fired at an elevation of 30°. If the muzzle velocity is 1000 ft. per 
second, determine the following : (a) equation of path ; (6) range ; (c) time of 
flight ; (d) position of the projectile after 2 sec; (e) highest point reached. 

V3 3000000 ' w 64 

( C ) 1^00 ; (d) aj = 1000 V3, y = 936 ; (e) ^^ ft. 

2. In problem 1 find the magnitude and direction of the velocity and its axial 
components after 20 seconds. 

Ans. v = 500 VS ft. per second ; v = — 140 ft. per second. 

3. Discuss the circumstances of the motion in problem 1 : (a) after the pro- 
jectile has passed over a horizontal distance * = 5000 ft., (6) when at an elevation 
of 1000 ft. 

4. A projectile moves subject to the equations x = at, y = bt — \ gt 2 . Discuss 
its motion fully. 

5. Show that a given gun will shoot three times as high when elevated at an 
angle of 60° as when fired at an angle of 30°, but will carry the same distance on a 
horizontal plane. 

6. The range is 300 ft. and the time of flight 5 sec. Find the initial velocity 
and the angle of elevation. What is the effect on the range of doubling the initial 
velocity ? Ans% Vo _ 100 ft< per seC0 nd ; sin a=$. 

7. (a) A boy can throw a stone 75 yd. on a level. How far from the base can 
he throw standing at the top of a cliff 150 ft. high ? (6) If the stone is thrown 
horizontally and strikes 450 ft. from the base, what is its initial velocity ? 

Ans. (a) 25v / 2lyd.; (6) 60 V~6 ft. per second. 

8. The wheels of an engine running at the rate of 40 mi. per hour encounter 
a drop of one quarter inch at the rail joint. How far from the joint will the wheels 
strike the lower rail ? Ans. - 1 - 1 v^ ft. 

9. Show that to strike an object at a distance x on the horizontal plane through 

the starting point, the elevation must be a or 90° — a where a = - sin- 1 -^- How 

2 V 

do the striking velocities compare in the two cases ? 

10. A bicyclist riding a wheel 28 in. in diameter notes that a piece of mud 
flying off the top of his wheel has a range of 12 ft. Find the angular velocity of 
the wheel and the cyclist's speed per hour. 

Ans. w = if± Vf rad. per second ; -V^-V f mi. per hour. 

11. A fountain sends out water horizontally in all directions from a central 
point a ft. high, with a velocity of c ft. per second. What is the shape of the water 
surface and the equation of a section made by the XT-plane ? 

Ans. x * + y* = ^^. 
9 



MOTION OF A PAETICLE IN A CONSTANT FIELD 147 

12. A drop of water flies off a grindstone just at the top. The radius of the 
stone is 2 ft., and it makes 1.5 revolutions per second. Find the velocity of the 
drop and the point at which it strikes the floor 7 ft. below the axis of revolution. 
What is the range if the drop flies off at a point such that the angle of elevation 
is 60° ? 

Ans. v = 6 VIO + tt 2 ft. per sec. ; x = | x ft. ; range = 9 ^^ ^ 2 + 3ttV27 ^ + 512 f 

2 ° 32 

13. "Where must the drop of water leave the grindstone in problem 12 in order 
to fall squarely on top of it ? In order to fall tangent to the opposite side ? 

Ans. cos a = , cos a = ■ — • 

9x 2 -8 9tt 2 

14. An emery wheel, 1 ft. in diameter, bursts into small particles when revolv- 
ing 100 times per second. Which particle will fly the farthest, and what is its 
initial velocity and range ? 

15. Show that the area of a level plane swept by a gun at a height h above the 
plane increases proportionally with h, being equal to A + 2hVirA where A is the 
area commanded when the gun is at the level of the plane. 

16. What must be the elevation a to strike an object 100 ft. above the horizon- 
tal plane and 5000 ft. distant, the initial velocity being 1200 ft. per second ? 

Ans. a is given by the equation 9 cos 2 a — 450 sin a cos a + 25 = 0. 

17. An engine can send a stream of water vertically 125 ft. How much of a 
vertical wall, distant 200 ft., can the engine wet ? Ans. 45 ft. 

18. Show that the area commanded by a gun on a hillside is an ellipse, with 
one focus at the gun. Find the area commanded by a gun which has a muzzle 
velocity of \ mi. per second, the slope of the hill being 10°. 

19. Determine the angle of projection so that the area included between the 
path and the horizontal plane is a maximum. Find the area. 

Ans. a = 60° ; area = -^ ^/3. 

20. Determine the elevation if the range on a given inclined plane is a 
maximum. 

Ans. Direction of projection must bisect the angle between the vertical 
and the inclined plane. 

21. Show that the range B of a projectile fired from a height h above a hori- 
zontal plane with velocity v at an angle a is given by 

2 v 2 (h + B tan a) = gB 2 sec 2 a. 

22. A heavy particle descends the outside of a circular arc whose plane is 
vertical. Prove that when it leaves the circle at some point Q to describe a 
parabola the circle is the circle of curvature of the parabola at Q. 

23. From a train moving at 60 mi. per hour a stone is dropped. The stone 
starts at a height of 8 ft. above the ground. Through what horizontal distance 
does the stone pass before it strikes the ground ? j lUS> 44 V2 ft. 

24. If the greatest range down an inclined plane be three times the greatest 
range up, show that the plane is inclined at 30° to the horizon. 



148 THEORETICAL MECHANICS 

25. Two balls are projected from the top of a tower, each with a velocity of 
50 ft. per second, the first at an elevation of 30°, and the second at an elevation of 
45°. They strike the ground at the same point. Find the height of the tower. 

Ans. (9 - 5 V3) -^9 = 26.5 ft. 
9 

26. The back lines of a tennis court are 78 ft. apart, and the service lines 
42 ft. The net is 3 ft. 3 in. high. Find the horizontal velocity of the ball (a) when 
it is returned from near the ground at one back line so as to graze the net and 
just strike the other back line ; (&) when it is served from a height of 8 feet, grazes 
the net, and strikes the service line. 

Ans. (a) 86.4 ft. per second ; (6) 170.64 ft. per second. 

27. The Norwegian ski jumping contests in February, 1904, took place on a 
snow slope at Holmenkollen 186 yards long. The competitors slid down § of the 
slope (which was in this part inclined 15° to the horizon) to a ledge, from which 
they took off for the jump. Below the ledge the steepness of the slope increased to 
24°. Supposing that the lip of the ledge was so curved as to give the jumper an 
elevation of 6° above the horizon at the take-off, find the speed at the ledge and the 
length of the leap. 

28. A particle is projected with velocity 2y/ag so that it just clears two walls 
of equal height a, which are at a distance of 2 a from each other. Find the time 

of passing between the walls. JE 

Ans. 2 v/-. 
y 9 

29. A gun is aimed directly at a target suspended to a balloon. Show that the 
bullet will strike the target if the latter is dropped at the instant the gun is fired. 

30. Show that the greatest range on an inclined plane through the point of 
projection is equal to the distance through which a particle could fall freely during 
its time of flight. 

31. Three bodies are projected simultaneously from the same point in the 
same horizontal plane, one vertically, another at an elevation of 30°, and the third 
horizontally. If their velocities be in the ratio 1:1: V3, show that they are always 
in a straight line. 

32. A heavy particle is placed very near the vertex of a smooth cycloid having 
its axis vertical and vertex upwards. Find where the particle runs off the curve 

and prove that it falls on the base of the cycloid at the distance ( - + V3 ) a f rom 

the center of the base, a being the radius of the generating circle. 

66. Constrained motion. On account of its fundamental prac- 
tical importance, we shall assume the constant field to be the gravi- 
tational field. If a particle falls along any smooth path under the 
action of gravity, we find the acquired velocity as usual by using 
the energy equation. 

If v and v are the velocities at A and B, respectively, the 
change in kinetic energy is 

\mv l — ^mv 2 . 



MOTION OF A PARTICLE IN A CONSTANT FIELD 149 



By Art. 52 the work done equals the force times the total dis- 
placement in the direction of the force, that is, equals nig(y — y), 
since the acting force is weight. Hence y 
we have 



mv, 



o ' 




from which we find 

(1) *=2g(y Q -y)+v*. 

If we set y — y '— height fallen = h, then (1) becomes 
. (V) t* 2 = 2 gh + V- 

The final velocity, therefore, depends upon the initial velocity 
and the height fallen, and is independent of the path. If v = 0, 
that is, if the body falls from rest, then 

(2) v 2 = 2gh or v = -\/2yh. 

This expression, V2#A, is called the speed due to a fall through 
the height h. 

The time of falling down any smooth curve is, however, not 
the same for all curves. Examples appear below. 

The intrinsic force equations are useful, and are readily 
written down, since the particle is acted upon by two forces only, 
weight and the normal pressure P n of the curve. Hence 



(3) 



n9 dv 

m — — = mv — = tangential component of weight, 
etc as 

v 2 
m — - = P n + normal component of weight. 
R 



67. Inclined plane. A particle constrained to move along a 
straight line oblique to the vertical is said to move along an in- 
clined plane. > The angle between the in- 
clined plane and a horizontal plane is called 
k the inclination. 




Smooth plane. The forces acting on the 
particle are weight and the pressure of the 
plane. Their resultant F must act along the plane. Taking the 
positive direction upwards along the inclined plane, we have 
from the figure 



a) 



F = — mg sin a, P n = mg cos a. 



150 THEORETICAL MECHANICS 

The particle therefore has an acceleration down the plane equal 
to g sin a. Consequently formulas (I) and (II) apply by setting 

(2) / = - g sin a. 

Rough plane. When a particle slides along a rough curve, the 
tangential component P t of the pressure of the path is called 
sliding friction. The following laws characterize this force: 

1. The direction of friction is opposite to the direction of 
motion. 

2. The magnitude of friction in any problem is directly pro- 
portional to the magnitude of the normal pressure. 

From the second law we have 

Friction = P t = /jl P n , 

where /jl is a constant called the coefficient of friction. 
Hence in the case of an inclined plane, 

Friction = /jl P n = /jl • mg cos a. 

Since friction is always opposed to the motion, for the result- 
ant acceleration the formulas obviously are 



(3) 



' / — — g (sin a + fi cos a), when the particle is moving up 

the plane. 
f = — g (sin a — fi cos a), when the particle is moving 

down the plane. 



In either case the acceleration is constant and (I) and (II) 
apply. 

The expressions (3) are made more compact by introducing 
the angle of friction. This is defined as an angle X whose tangent 

is the coefficient of friction; that is, 

(4) /jl = tan X. 

To see the significance of X, consider 
a particle at rest upon a rough plane. If 
now the plane be tipped so that the in- 
clination increases, the particle will even- 
tually move. The inclination when the particle is on the point of 
moving equals the angle of friction. For at this instant the re- 




MOTION OF A PARTICLE IN A CONSTANT FIELD 151 



sistance of the friction equals the component of weight down the 
plane; that is, 

j mg sin a = jx mg cos a; or 
^ ) \ tan a = a. . \ a = X. 



(6) 



Substituting //, = tan X in (3) gives 

f = — g sec X sin (a + X), when the particle is moving 

up the plane. 

f = — g sec X sin (a — X), when the particle is moving 

down the plane. 

The numerical value of the coefficient of friction depends upon 
the character of the substances in contact, and is determined by 
experiment. The value of ^ is slightly less when the particle is in 
motion than when it is at rest but on the point of moving. 



ILLUSTRATIVE EXAMPLES 

1. A particle is projected up an inclined plane with the velocity v > How far 
will it ascend ? 

Solution. Since the motion is resisted by a constant force, we have, by the 



2f 





energy equation, \ mvo 2 = mfs, where s is the distance required, 
where f=g sin a or g (sin a + fx cos a) according as the q 

plane is smooth or rough. 

2. A heavy particle starts from rest at the top of an 
inclined plane. Eequired the locus of the foot of the 
plane if the time of descent is constant and independent 
of the inclination. 

Solution. Smooth Plane. By (I), 

s = \ g sin a t 2 = \ gt 2 sin a. 

In this equation s and a are 
variable. If is the starting point, it is easily seen that 
the required locus is a circle having as the highest point 
and a diameter OH = i gt 2 . For s = OQ = OH sin OHQ 
= OH sin a = \ gt 2 sin a. The result may also be stated 
thus : The time of descent along all smooth chords of 
a vertical circle drawn from its highest point is the 
same. 

Bough Plane. In this case from (I) and (6) Art. 67, 

7 sec X sin (a — \)t 2 = ± gt 2 sec \ sin (a — X). 

As before, let the vertical line OH- \ gt 2 . Lay off the angle HOD = X, and 
construct the arc of a circle whose center lies on OB passing through O and H. 



I 




152 THEOKETICAL MECHANICS 

This arc is the required locus when the particle descends to the left of OH. For 
OD = OH see X = \gt 2 sec X. Also 

s = OQ = OD ■ sin ODQ = OD sin (a - X) = J gt 2 sec X sin (a - X). 

If the particle moves to the right, the locus is a 
corresponding equal arc. 

Line of quickest descent from a point to a given 
curve. Given a curve C and a point 0. If a vertical 
circle tangent to C and having for the upper ex- 
tremity of a vertical diameter is drawn, then Q is a 
line of quickest descent along all smooth straight lines 
from to C. For along OS the time is greater than 
down OQ, since the time along OP equals that 
along OQ. 

PROBLEMS 

1. A smooth plane has an inclination of 30°. With what velocity must a par- 
ticle be projected up the plane, the length of which is 48.3 ft., that it may just reach 
the top ? What must be the initial velocity to reach the top in 1 sec? (g = 32.2.) 

Ans. v = 16.1 V6 ft. per second; 
v = 53.35 ft. per second. 

2. A particle falls from rest down a given inclined plane. Compare the times 
of descending the first and second halves. Ans. 1 : V2 — 1. 

3. Along what chord of a circle must a particle fall in order to gain half the 
velocity which it acquires in falling through the vertical diameter ? 

Ans. Chord inclined 60° to vertical. 

4. A particle is projected up a smooth plane which has an inclination of 3 in 5 
with a velocity of 40 ft. per second. In what time will it come to rest and how far 
up the plane will it go ? Ans. f f sec. ; J-f * ft. 

5. A weight of 10 lb. falls vertically and draws a 15-lb. weight up a smooth 
plane having an inclination of 30°. What is the acceleration, pull on the string, 
and space fallen through in 10 sec. ? Ans. f = ^g: T = 9 lb. ; space = 5g. 

6. A railway train is running at the rate of 30 mi. per hour up a grade of 
1 in 50. The coupling breaks, cutting loose part of the train. How long will the 
detached part continue up the grade, friction being neglected ? What is its position 
with respect to the point where the break occurred and what is the direction and 
velocity of its motion after 2 minutes ? Ans. ^f 1 sec. ; if- 4 ft. per second downhill. 

7. With what velocity must a particle be projected down an inclined plane of 
length I so that the time of descent shall be the same as that for a free fall through 

the height of the plane ? A I — h sin a 

° r Ans. v = g . 

^/2gh 

8. What is the value of g if a given mass descending vertically draws an equal 
mass up an incline of 30° a distance of 25 ft. in 2.5 sec? Ans. g = 32. 

9. Find the position of a point on the circumference of a circle such that the 
time of descent from it to the center shall be the same as the time of descent from it 
to the lowest point of the circle. 



MOTION OF A PARTICLE IN A CONSTANT FIELD 153 

10. A particle slides down a smooth plane inclined at an angle of 45° and then 

drops into free space. (1) If the particle has a velocity of 20 ft. per second when it 

leaves the plane, find the equation of its path. (2) Where will it strike a horizontal 

plane 100 ft. below ? (3) What are the axial components of the velocity when 

it strikes this plane ? (4) At what point will it cut a vertical line 45 ft. distant 

from the plane ? .- ,- ft v '2 

Arts. (3) v x = 10 V2, v y = 10 V2 + ^^ ; (4) 207 ft. below. 

5 

11. A train of 100 T. starts on an up grade of 1 in 50 with a speed of 20 mi. 
per hour. It is stopped by gravity and the resistance of the brakes in 4 seconds. 
(1) What is the coefficient /* of the resisting forces ? (2) What is the velocity at 
the end of one second ? Ans. (1) /x = .21. 

12. Show that the times of descent down all radii of curvature of the cycloid 

are equal. /j^ 

Ans. T = \l—. 
X 9 

13. A heavy particle starts from rest at the top of an inclined plane. Ee- 
quired the locus of the foot of the plane if the speed at the foot is constant and 
independent of the inclination. AnSt A straight line. 

14. Give a construction for finding the line of quickest descent from a fixed 
point to a circle in the same vertical plane. 

15. A body begins to slide down an inclined plane from the top, and at the 
same, instant another body is projected upwards from the foot of the plane with 
such a velocity that the bodies meet in the middle of the plane. Find the velocity 
of projection and determine the velocity of each body when they meet. 

Ans. V2 gh • 0, and V2 gh, where h = vertical height of the plane. 

16. A parabola is placed with its axis vertical and vertex upwards. Find the 
chord of quickest descent from the focus to the curve. 

Ans. The chord makes an angle of 60° with the vertical. 

17. Through what chord of a vertical circle drawn from the bottom of the 

vertical diameter must a body descend so as to acquire a velocity equal to ~ th part 
of the velocity acquired in falling down the vertical diameter ? 

Ans. If 6 denote the angle between the required chord and the vertical diam- 
eter, cos 6 = - . 
n 

18. A heavy particle is projected up a smooth inclined plane with a velocity of 
36 ft. per second. The inclination of the plane is 30° and its vertical height is 20 ft. 
It projects into space at the top of the plane. Determine (a) the time in ascend- 
ing the plane, (&) the velocity at the top, (c) the equations of the free path. 

19. A body is projected up a rough inclined plane with the velocity which 
would be acquired in falling freely through 12 ft. , and just reaches the top of the 
plane. If the inclination of the plane is 60° and the angle of friction is 30°, find 
the height of the plane. ^4^ s . 9 ft. 

20. A body is projected up a rough inclined plane with the velocity 2 g. If the 
inclination of the plane is 30° and the angle of friction is 15°, find the distance 
along the plane which the body will move. Ans. tf(V3 + 1) ft. 



154 



THEORETICAL MECHANICS 



21. A body is projected up a rough inclined plane, inclination a, and the angle 
of friction is X. If m be the time of ascending and n the time of descending, show 
that /m\ 2 _ sin (a — X) 

\n) sin (a + X) ' 

22. A particle descends an inclined plane. If the upper portion be smooth and 
the lower rough, coefficient of friction being fx, and if the smooth length be to the 
rough length asp : g, show that the particle will just come to rest at the foot of the 



plane if [x 



P + 



tan a, where a is the inclination of the plane. 



23. Two rough planes, coefficient of friction = /i, inclined respectively at angles 
a and a' to the horizon, are placed back to back as shown in the figure. Two 

masses, m and m', are placed upon them, being con- 
O^v nected by a string passing over a pulley at O. (a) If 

m = m', find the limit of the difference a — a', if the 
acceleration is zero. (6) If a = «', find the limit 
of the difference m — m' if the acceleration is zero. 

24. Particles are sliding down a number of wires which meet in a point, all 
having started from rest simultaneously at this point. Prove that at any instant 
their velocities are in the same ratio as the distances they have traversed. 




68. Motion on a smooth circle. Simple pendulum. A heavy 
particle is suspended from a fixed point by an inextensible 
thread, and swings under the action 
of its weight in a vertical circle. 
Discuss the motion, neglecting the 
weight of the thread and the in- 
fluence of the atmosphere. 
Let I = length of thread, 

A be the initial position 
(t = 0) at rest, 
P be the position after time t, 
s = arc AP. 
Then, from the figure, 
(1) * = Z(a-0). 

Resolving the impressed force mg into tangential and normal 
components, we have the intrinsic force equations : 

d 2 s 




(2) 



(3) 






dt* 



f normal 



mv 



— = < impressed 
[force 



P t = mg sin 6, 



normal 
pressure 



— mg coi 6 — T, 



MOTION OF A PARTICLE IN A CONSTANT FIELD 155 

where T = tension in the thread, that is, the pull of the particle on 
the thread. 

The work done by weight when the particle descends from A 

to P ls mg • MN = mgl(cos - cos a) . 

Hence, the energy equation gives 

(4) ^ v 2 = gl(cos — cos a). 

The tension in the thread at any instant is given by eliminating 
v 2 from (3) and (4), that is, 

(5) T = mg(2 cos a — 3 cos 0) . 

The circumstances of the motion are known if is known as a 
function of the time. 

From (2) g=y S in<?, 

and from (1) S=~*fr 

Hence, the differential equation for the determination of is 

Small amplitudes. The angle a, which is the maximum value 
of 0, is called the amplitude of the motion. A simple solution of 
(6) results if a and consequently are small. A close approxi- 
mation is then found by assuming sin = 0. Equation (6) then 
becomes 

(T) g + f,-0. 

The general solution of (7) is (see 71, Chap. XIV) 

(8) = c x cos Y~ 7 1 + c 2 sin \/| £. 

The constants of integration c x and c 2 are to be determined by 
the initial conditions. 

When t = 0, = «, — = 0. 
dt 



Hence, c x = a, and c 2 = 0, and (8) becomes 
(9) 0=acosyjft. 



156 THEORETICAL MECHANICS 

From (9), the motion is periodic, the time for a complete oscil- 
lation being 
(VI) JP=2ir fl. 

For any given position on the earth's surface g is a constant. 
Hence (VI) shows that the period depends only upon the length of 
the pendulum. It must be remembered that this formula holds 
only when the amplitude is so small that the substitution of the 
angle for its sine is within the limit of error. 

Any amplitude. To obtain an expression for the period which 

is valid for any value of the amplitude, we proceed as follows. 

d6 
Multiplying equation (6) by — , we may integrate each term, ob- 

. • ■ Cto 

taming 

1 fd6\ 2 g a 

- —-) — f COS0= c. 
2\dtJ I 

Since — = when = a, e = — ^cos a, and we may write the 
dt I 

equation in the form 



© 2 = 2 !< cos *- cosa >- 

It is convenient to set in this equation cos 6 = 1 — 2 sin 2 J0, 
cos a— 1 — 2 sin 2 !- a. Extracting the square root of both sides, 
and taking the negative sign with the radical, since decreases 
as t increases, we obtain 

d6 



^--9-kI 



dt " *l ^si^J^-sin 2 ^- 

The time required for the particle to move from the highest 
point (6 = a ) to the lowest (0 = 0), which is one fourth the period, 
is obtained by integration, namety, 

dO 



9 J Vsin 2 l«-sin 2 l(9 



or 



a 

(10) P = 2yf- f a 



gJ Q Vsin 2 1 a — sin 2 \ 6 



MOTION OF A PARTICLE IK A CONSTANT FIELD 157 



The integral in (10) is transformed into a known form by 

setting 

sin ^ a = k, sin ^ 6 = k sin $, 

where <f> is the new variable. Differentiating, we get 

J cos J 6d6 = k cos (f>d<f>. 

nr\ _ 2& cos cf)d(f> _ 2 k cos $>d<$> 

cos 1 6 " Vl-^sin 2 ^' 

Since if = 0, = 0; = «, $ = — , we obtain for (10) the form 



(11) 



P=4 



if 



<ty 



Vl — k 2 sin 2 (/> 



(& = sin -|- a) 



The integral involved here is known as the complete elliptic 
integral of the first species and is denoted by K. The integral 
evidently is not independent of a. Indicating this dependence 
by a subscript, then 

(12) P. = ^K4- a 

Values of K may be found tabulated (p. 117) in Peirce's A 
Short Table of Integrals (Ginn and Company). 

A few values of K are set down here in order to make clear 
the dependence of the exact period P a upon the amplitudes. 
Comparing (12) and (VI), we have 



(13) £-*£ 



or, 



P a -P K-lir 



Pa 



K 



Remembering that P is an approximate 
value of the period for small amplitudes, (13) 
gives the percentage of error. For example, 
when a = 10° this is 



1.5738-1.5708 
1.5738 



^ = about * of 1%. 



An approximate form of P a closer than 
the value of P is found by writing (11) in the form 

(14) P a = 4^/- P(l - & 2 sin 2 4>) _l #. 

9 Jo 



a 


K 


0° 


\ir 


QO 

+4 


1.5709 


4° 


1.5713 


6° 


1.5719 


8° 


1.5727 


10° 


1.5738 



158 THEORETICAL MECHANICS 

Using the Binomial Theorem (1, Chap. XIV) then 
(1 - k 2 sin 2 </>)"* = 1 + \W sin 2 c/> - etc. 
Substituting and integrating gives 

P a = 4 yll f'l+ 7 Lk2 + terms in k\ etc.\ 
*gV2 8 J 

1 1 a? 

Since k = sin -a = -a— — + •••, if we neglect all powers of a 

2 2 48 

higher than the second, we obtain 

as) ^= 2 n/^+S= p ( 1+ i9- 

This formula is useful in practice. P a may be observed and 
the value of P then determined. 

69. Motion on a smooth cycloid. The discussion of Art. 68 
demonstrates that the period of oscillation of a heavy particle on 

a vertical circle depends upon 
the amplitude. The question 
T arises : Does any curve exist upon 
\ /A. which the period of the oscilla- 

l — ..^Cy\ tions is the same for all ampli- 
t ^^y\ M — \ — tudesf The cycloid possesses 
~x; ' this property, as will now be 

shown. 
The equation of the cycloid of the figure is (Calculus, p. 281), 

y 



(1) x = a arc vers ~ + V2 ay — y 2 . 

We shall show that the time of descent to the lowest point 
from rest at A is the same for all positions of A. 

The velocity at P is found by the energy equation to be 



(2) v 2 =2g(y -y) or — = - V2# Vy - y, 

if s is measured from the lowest point. Hence the time from A 
to is given by 



(3) t 



= -—{ 

V2 a Jy 



gjy« Vy -y 



MOTION OF A PARTICLE IN A CONSTANT FIELD 159 

We must now express s in terms of y. To do this, differentiate 
(1), which gives 



dx _ 2 a — y _ 12 a 



dy -V2ay-y 2 V 

Equation (3) now becomes, by substitution, 

(5) *=-Vf/-^=^. 

^e/2/o vy y-y 2 g 

This result depends only upon the height of the cycloid and is 
therefore the same for all positions of A. 

A second demonstration is important. Differentiating (2) 
with respect to s, we obtain 

(6) 2v^=-2g^- 

v J ds y ds 

Hence by the intrinsic force equations, 

as 

Let s be measured from the position of equilibrium 0. Then we 
have, from (4), 

(7) s=OP = C V ^1± dy = 2 sj2a~y. .-. s 2 = 8 ay. 

Jo -Vy 

Differentiating this, we obtain -£ = 

ds 4 a 

Hence from (6), we have, 

The general solution is (see 71, Chap. XIV) 
8 = c x cos ~\/- t-\-c 2 sin ^Y~^ * 

— (A —it 

The motion is therefore a harmonic curvilinear oscillation about 
with the period 4 7r\'-« 

Conversely, it may be shown that the cycloid is the path of a 
heavy particle which performs oscillations of equal periods on a 



160 THEOEETICAL MECHANICS 

smooth curve. The oscillations are said to be tautochronous 
(equal times) and the cycloid is the only tautochronous curve 
when the impressed force is gravity. 

70. Seconds pendulum. If the period of a simple pendulum 
for small amplitudes at a given locality on the earth's surface 
is two seconds, the pendulum is called a seconds pendulum for 
that place. Since g varies along the earth's surface, the length of 
such a pendulum is necessarily variable, but is about 39.11 inches. 

For points without the earth, gravity varies inversely as the 
square of the distance from the earth's center. Hence if g f is the 
intensity of gravity at a height x above the earth's surface, we 
shall have 

n\ 9' - R2 '- C > R2 

g~(x'+Ry* ° V 9 ~{R + xy' 

where R = radius of the earth. Hence if P x denotes the period at 
the height a?, then 

(2) P, = 2^=2.VI(^)=P(1 + | 

This formula gives the relation between the periods of the 
same pendulum at the earth's surface and at any height x. 

For points in the interior of the earth, gravity varies directly 
as the distance from the earth's center. Hence, if g f is the in- 
tensity of gravity at the distance y below the surface of the earth, 
we shall have 

(3) * = *!?*• 



PROBLEMS 

1. If 39.11 in. be taken as the length of a seconds pendulum, that is, a 
pendulum which makes one full swing in one second, what is the length of the 
pendulum which vibrates 25 times per minute ? whose period is f ir ? 

2. In what time will a pendulum vibrate whose length is double that of a 
seconds pendulum ? 

3. A pendulum which beats seconds in London requires to be shortened by- 
one thousandth of its length if it is to keep time in New York. Compare the values 
of gravity at London and New York. 

4. What is the length of the seconds pendulum where g = 980 cm. per sec- 
ond per second ? 



MOTION OF A PARTICLE IN A CONSTANT FIELD 161 

5. Prove that a seconds pendulum if brought to the height x mi. will lose 
about 22 x sec. per day, the radius of the earth being taken as 4000 mi. 

6. Find the height at which a pendulum 60 cm. long will beat seconds, taking 
the radius of the earth as 4000 mi. 

7. The mass of a pendulum bob is 100 gm., and the string is 1 m. long. 
What is the kinetic energy when the string makes an angle of 30° with the vertical 
if the bob is dropped from a horizontal position ? Ans 8.5(10) 6 ergs. 

8. A body whose mass is 1 lb. is suspended from a fixed point by a string 12 
ft. long. The string is swung to a position 60° from the vertical and the body re- 
leased. Determine the velocity when the body is in its lowest position ; also when 
2 ft. above its lowest position. 

9. A clock gains 3 min. per day. How much should the bob be screwed up 
or down ? Ans , Down by _i^ f its length. 

10. Find the time, to four decimal places, of a half vibration of a pendulum 
1 m. long at a place where g — 980.8 cm. per second per second. 

11. The seconds pendulum loses 12 sec. per day when carried to a mountain top. 
How high is the mountain ? Ans About 2900 ft. 

12. Find the time of vibration of a seconds pendulum placed in a mine 1.5 
mi. deep. 

13. Compare g at two places where the rates of the same pendulum differ by 
5 vibrations per hour. 

14. A string r ft. long has a mass m attached to the lower end and acts as a 
simple pendulum. Find the point in the arc where the pull on the string is the 
same as where the pendulum is at rest. 

Ans. y = 1 7i, where h is the height from which the pendulum has fallen. 

15. A heavy particle oscillates in a complete cycloid from cusp to cusp. 
Prove the following properties : 

(1) The velocity at any point P 
equals the velocity at the lowest point 
resolved along the tangent at P. 

(2) The time of description of any 
arc OP is proportional to the angle 

OAQ = r. In fact r = Ai— • t. 

(3) If the particle is regarded as rigidly attached to the generating circle, then 
the center of the latter moves with constant speed. 

(4) The pressure on the curve equals twice the normal component of weight. 

(5) The acceleration of the particle is equal to g and is directed towards the 
center of the generating circle. 

Hint. Use the equations x = a{6 + sin 6), y = a(l — cos 0), the properties in- 
dicated in the figure, and the relation R = 2 AQ (P = radius of curvature). 

16. In the motion of a particle down a cycloid, prove that the vertical velocity 
is greatest when it has completed half its vertical descent. 



Y 


A 




^ / 









/ X 



162 THEORETICAL MECHANICS 

17. When a particle falls from the highest to the lowest point of a cycloid, 
show that when it has described half of the path, f of the time has elapsed and it 



18. The bob of a pendulum which is hung close to the face of a vertical cliff is 
attracted by the cliff with a force which would produce an acceleration / in the bob. 



I' 2 
Show that the time of a complete oscillation is 2 ir\l , where I is the length of 

V+/ 2 

the pendulum, and find the center of the arc described by the bob. 

19. A railway train is moving uniformly along a curve at the rate of 60 mi. per 
hour, and in one of the carriages a pendulum, which would ordinarily beat seconds, 
is observed to oscillate 121 times in 2 min. Show that the radius of the curve is 
very nearly a quarter of a mile. 



CHAPTER VII 

CENTRAL FORCES 

71. Central field of force. A field of force is called a central 
field if the direction of the acceleration at every point of the field 
passes through a fixed point called the center of force. The accel- 
eration may be directed towards the center of force or from it ; that 
is, the force may be attractive or repulsive. The magnitude of 
the acceleration may vary according to any given law. In the 
general case it may depend upon the direction and the distance 
from the center and also upon the time. In many practical prob- 
lems, however, the magnitude of the acceleration depends only 
upon the distance from the center, and we shall confine our atten- 
tion to this case. The term central force, therefore, as used here 
applies to central fields in which the magnitude of the force de- 
pends only upon the distance from the center of force. Let the 
origin of coordinates be taken at the center of force and P be 
the position of a material particle subject to the force of the field. 
Denote the distance OP, which is called the radius vector, by p. 
Then the magnitude of the acceleration exerted upon the particle 
at P is a function of p, 

and its direction is along the line OP, in the positive sense if 
the force is repulsive, and in the negative sense if the force is 
attractive. 

The path of the particle is called the orbit. If the initial 
velocity is along the line OP, the orbit is a straight line, since 
the force has no component tending to draw the particle out of 
the line. If the direction of the initial velocity is oblique to the 
line OP, the orbit is a plane curve, since the force has no com- 
ponent tending to draw the particle out of the plane determined 
by the direction of the initial velocity and the line OP. The 
orbit is, then, necessarily a plane curve. 

163 




164 THEORETICAL MECHANICS 

Since the acceleration is directed towards the concave side of 
the path we may draw the conclusions : (1) if the orbit is con- 
cave towards the center of force, then the force is attractive ; (2) 
if the force is attractive, the orbit is concave towards the center of 
force. 

72. Areal velocity. When a point moves along a curve, its 
radius vector is said to generate area. Thus, if the moving point 
describes the curve c in the figure from the 
point A to the point B, its radius vector gen- 
erates the area AOB. The time-rate at which 
the radius vector generates area, that is, the 
derivative of the area with respect to the time, 
is called the areal velocity of the moving 
point. By Calculus, p. 377, the differential of area in polar coor- 
dinates is 

Hence the areal velocity is given in terms of the angular velocity 
by the relation 

^ 1X dA 1 9 d6 1 o 

(1) -dt = -/Tt = '/'°- 

To derive an expression for the areal velocity in terms of the 
rectangular components of velocity, we proceed as follows. The 
rectangular and polar coordinates of a point are connected by 
the relations 

f X = p cos 0, 
^ 1 y = p sin 6. 



Differentiating with respect to £, 
(3) 



r dx dp a . Q d0 

— - = -J- cos 6 — p sin 6 — , 
dt dt dt 



dy dp . * , Add 

-2- = -J- sin 6 + p cos u — . 
L dt dt dt 

Multiplying the first of equations (3) by sin 0, the second by 
cos 0, and subtracting, we get 

/i dy • /) dx dS 
cos 6 -i?- — sin v — - = p — - ■ 
dt dt dt 



CENTKAL FOKCES 165 

Multiplying by | p and taking account of (2), we have finally 

> . N tf ^4. 1 / dy dx\ 1 , • N 

This equation expresses the areal velocity in terms of the 
rectangular coordinates and their derivatives with respect to the 
time. 

73. Law of areas for central forces. Since a central force acts 
in the direction of the radius vector, its component perpendicular 
to the radius vector is zero. Hence, using polar coordinates, the 
differential equations of motion, Art. 51^ of a particle of mass 
m are 



a) 



[dfi l \dtJ . 



F 9 = mf, 



-$>©-*-* 



where /denotes the acceleration. The second equation gives by 
integration 

(2) P f t = n, 

where h is a constant. 

Comparing (2) with (1), Art. 72, we have 

(3) 2 ^ =A - 

Hence the 

Theorem.* In the motion of a particle subject to any central 
force the areal velocity is constant. 

The constant h, which is twice the areal velocity, is called the 
constant of areas. 

Integrating equation (3) between the limits t = t 1 and t = t v 
we have , 

that is, the area generated in any interval of time t 2 — t x is pro- 
portional to the length of the interval. In other words, the 
radius vector sweeps over equal areas in equal intervals of time. 

* Since the law of variation of the force, that is, the function F, does not enter in the 
derivation of (3), it is evident that this theorem and the two following theorems in this 
article hold also for the general central field of force, that is, when F may depend upon 
p, 6, and t. 



166 THEORETICAL MECHANICS 

From (2), 

Hence the 

Theorem. In the motion of a particle subject to any central 
force, the angular velocity is inversely proportional to the square of 
the distance from the center of force. 

The speed of the particle is 

ds _ ds d6 
dt~d6~di' 

or, substituting the value of — from (4), 

017/ 

^rx ds _ds h 

c ; Tt~d6j*' 

Let p denote the perpendicular distance from the origin to the 
tangent to the orbit. Then from the figure, 

I = sin yjr = 4- (Calculus, p. 98). 
p ds 




dd 



Hence — = (—. 
d6 p 



Substituting this value in (5), we obtain 
ds h 

Hence the dt ~P 

Theorem. In the motion of a particle subject to any central 
force the speed is inversely proportional to the perpendicular distance 
from the center of force to the tangent to the orbit. 

74. Converse of the theorem of areas. Suppose a particle 
moves in a plane in such a manner that its radius vector generates 
equal areas in equal intervals of time, that is, its areal velocity is 
constant. Then j \ 

and o^=2h. 

r dt 



CENTRAL FORCES 167 

Differentiating with respect to t, 

But the first member of (1) is p times the component of accel- 
eration perpendicular to the radius vector ((1), Art. 73). There- 
fore the total acceleration is in the direction of the radius vector, 
that is, the direction of the acceleration passes always through the 
fixed point which is the origin of coordinates. Hence the 

Theorem. If a 'particle moves in a plane in such a manner 
that its areal velocity with respect to a fixed point in the plane is 
constant, then the particle is subject to the action of a central field 
of force with the point as center. 

75. The energy equation. The energy equation in polar coor- 
dinates is, Art. 62, 

W= £(•» - V) = r^F.dp+Fepdd. 

For central forces F =Q, and, under the assumption that the 
magnitude of the force depends only on the distance, we may 
write, 

F P =F(p), 

and the energy equation becomes 

(1) ^(j> 2 -«h=rF(p-)a P . 

A Jpo 

Let the function U(p) be defined by the equation, 

-if= + F, 

dp 
Then - U= C ? Fdp. 

espo 

Equation (1) may now be written 

(2) . f« 2 +^(p)=fV+^(Po)- 

The function U is called the potential function * and the value 
of U for any given value of p is called the potential energy of the 

* The subject of the potential function and potential energy is treated in Chapter X. 



168 



THEORETICAL MECHANICS 



moving particle. The kinetic energy is J mv 2 , and since the second 
member of equation (2) is constant, we have the 

Theorem. The sum of the kinetic energy and potential energy 
of a particle free to move in a central field of force is constant. 

Equation (2) is called the vis viva equation. The constant 
value of the second member depends upon the problem, that is, 
upon the initial conditions. When this has been determined, 
equation (2) defines v 2 (the square of the speed) in terms of U 
which depends upon p alone. In physical problems U is a single- 
valued function of /?, and from (2) the speed is uniquely 
determined if the distance from the origin is known. Hence the 

Theorem. In any given problem of the motion of a particle in 
a central field of force, the speed depends only upon the distance 
from the center of force. 

As examples of the application of the preceding theorem consider the following : 

(1) One end of an elastic string is fixed at the point 0. To the other end is 
attached a particle which moves under the action of the elasticity of the string 

(neglecting the friction of the air). Motion is 
begun by projecting the particle from a given point 
(/jo? #o) with a given speed vq. These facts deter- 
mine the constant value c of the second member of 

equation (2), namely c = ^L + U(p ). Then if a 

particle passes through a point A in any direction 
we can determine its speed vi if we know the dis- 
tance OA = pi. Furthermore, if the particle crosses 

the circle about with radius OA at any point as 

B, C, or D in any direction, its speed is the same 

as the speed at A, namely v\. 

(2) If a small planet or comet revolves in an 
ellipse about the sun under the sun's attraction, 
which is inversely proportional to the square of the 
distance from its center, the speed of the planet 

at the distance 8A when approaching the sun is the same as the speed at the dis- 
tance 8 A' — SA when receding from the sun. 

76. Circular orbits. The problem of determining the motion 
of a particle in a central field of force demands the solution of 
the differential equations of motion; 
~d 2 P fd6\ 2 ' 





(1) 






dt 2 p \dtJ 



m 



2 ddX 

dt\ p dt 



= F(p)=mf(p').. 



= 0. 



CENTRAL FORCES 169 

Integrating the second equation we have, 
(2) P f r k, 

where h is a constant of integration. Let us impose the condition 
that the particle shall move around the center of force in a circle 
of radius a. Then p = a and from (2), we get for the angular 

velocity, 

Substituting this value in the first of the equations (1) gives, 

sinoe S = 0, » ff . 

whence, 



(4) h=-V-a s f(a~). 

The value of h thus determined is real if f(a) is negative, 
that is, if the force is attractive. 
By integration of (3) we obtain 

(5) = \t + c=yE2Mt + c. 

a z * a 

Hence a particular solution of the differential equations (1) is 

(6) 

t + c. 

The constant c is determined if we know the position of the 
particle at any given time. For example, if = Q when t = 0, we 
find c = O . 

Theorem. In an attractive central field of force a particle 
may move around the center of force in a circle of given radius a. 

The angular velocity co is constant and equal to a/ — ^ K The 
speed is equal to aco. 

Illustrative Example. If the acceleration in a central field is towards the 
center of force and proportional to the distance, the time of describing a circular 
orbit is independent of the radius. 




170 THEORETICAL MECHANICS 

Solution. The acceleration is given by 

f = -k*r, 
where k is a constant. 

Hence the angular velocity in a circular orbit of radius a is 

m = 

a 

The constant angular velocity w = k is the angle (measured in radians) turned 
through by the radius vector in one unit of time. Hence the time required to de- 

2 7T 

scribe the complete circle is - — units. The time required for the particle to move 

k 

completely around its orbit is called the period. The period is constant and equal 

to — . 
k 

77. Differential equation of the orbit. To find the equation of 
the orbit of a particle in a central field of force we may integrate 
the differential equations of motion and then eliminate t. An- 
other method which leads to important results is to first eliminate 
t, obtaining a differential equation involving p and (or x and ?/), 
the integration of which furnishes the equation of the orbit. In 
this process it is convenient to use, instead of the radius vector 

/o, its reciprocal u = -. Then 
P 

,.. ^ dp _ 1 du _ 1 du d6 

^' dt~ ~^~di~~~yfi~de'di' 



But from (2), Art. 73, 

l_d£ 
u 2 dt 



09 ^=*. 



Hence * = - **£. 

dt dd 

Differentiating with respect to t and taking account of (2), 
r o\ d 2 p _ id fdu\ yd 2 ud0_ l2 2 d 2 u 

Substituting the value of —£ from (3) and the value of — 

dt dt 

from (2) in the first of the differential equations of motion [(1), 

Art. 76], we have 



m 



[- W S;->M3- 



CENTRAL FORCES 171 

Or 



«, "'© + •)- J 



Equation (4) is important for the solution of two problems: 

(1) Given the orbit, to determine the law of central force. 

(2) Given the law of central force, to determine the orbit. 
The solution of the first problem is usually quite simple. From 

the equation of the orbit we know u in terms of 0, u = </>(#). 
Hence we may compute the first member of (4), 

(5) iW)[fW + W)] = --- 

m 

Under the assumption that the law of force shall depend only 
on the distance, we may find 6 from the equation of the orbit and 
substitute its value in (5), obtaining F in terms of p and the con- 
stant of areas h. If h is known, the force is uniquely determined. 

One exceptional case must be mentioned. The process fails if 

the orbit is a circle about the center of force. If u = - and h is 

a 

given, equation (4) furnishes a value for the intensity of the force 
at the distance a from the center, but does not prescribe a law 
governing the intensity of the force at any other distance. It 
was shown in the preceding article that a circular orbit about the 
center is possible for any attractive central force. 

Illustrative Example. Determine the law of central force if the orbit is the 

circle p = 2 a cos 6. 

a 7 .. 1 sec 6 

Solution. u= — = , 

p 2 a 

du _ sec 6 tan 6 

dd~ 2 a 

d^u _ sec 3 6 + sec d tan 2 6 _ sec 6 (2 sec 2 6—1) 
dd 2 ~ 2 a 2 a 

Hence — = w(8 a 2 w 2 - 1). 

Applying (4), we have 

_ Z = ft2 M 2[ M (8 a 2 w 2 _ i) + W ] _ 8 aWu 5 . 
m 

Since u = - , the final expression for the force is 
P 

„ 8 aWm 

F= — 7~- 

The force is attractive and proportional to the fifth power of the distance. 
Hence the 



172 THEOEETICAL MECHANICS 

Theorem. If a particle describes a circle under the action of a center of force 
on the circumference, the force is attractive and varies inversely as the fifth power 
of the distance. 

PROBLEMS 

1. Assuming that the planets move around the sun in circles, prove Kepler's 
Harmonic Law, which states that the squares of the periods are proportional to the 
cubes of the distances. 

2. A particle describes a circular orbit with angular velocity w about a center 

/ Tc 2 tn\ 

of force which is inversely proportional to the distance ( F = ) • Deter- 
mine the radius of the circle. . ' k 

Ans. a = -. 

3. A particle describes an ellipse with the center of force at one focus. Show 
that the force is inversely proportional to the square of the distance. 

4. A particle describes an ellipse with the center of force at its center. Show 
that the force is proportional to the distance. 

Suggestion. The equation of a conic section with center at the origin is 

p 2 = » Theni^-^ 1 -* 2 ),. 

l-e 2 cos 2 6 4 

For an ellipse 1 — e 2 > and the force is attractive. 

For an hyperbola 1 — e 2 < and the force is repulsive. 

5. The orbit is an hyperbola with the center of force at the right-hand focus. 
Show that if the particle moves (a) on the right-hand branch of the curve, the force 
is attractive and inversely proportional to the square of the distance ; (5) on the 
left-hand branch, the force is repulsive and inversely proportional to the square of 
the distance. 

6. Find the central force under which a particle may describe the orbit given. 

" (a) the reciprocal spiral pd = a. Ans. F = ~ mi . 

P s 



(6) the logarithmic spiral p = e< Ans. F = ~ mh2 ( a * + 1 ) . 

(c) the lituus p 2 = a 2 . Ans. F=-mh 2 [-~ -£-\ • 

(d) the lemniscate p 2 = a 2 cos 2 0. Ans. F=- 8 h<1 ^ m • 

P' 

(e) thecardioidp = a(l + cos0). Ans. J=- 3gfa . 

(/) the limacon p = b - a cos 0. Ans. F = - mh 2 ( 2 ( a2 ~ b ^ + ^j\ - 

Cg) the four-leaved rose p = a cos 2 0. Ans. F = — mh 2 ( — 

V P 5 P s 

(h) the three-leaved rose p = a cos 3 0. Ans. F = — mh 2 ( — ^ 

V P 5 P 

(0 the rose p = a cos n0. Ans. F = -mh 2 (^^-^=^-\ 

\ P b P 3 / 



CENTRAL FORCES 173 

7. Find the law of central force under which a particle may describe the curve 
whose equation is 

p k = a cos kd+b, 

where a, 6, and k are constants. 

\ p 2k+3 T p k+Z J 

The curve in problem 7 includes many of the common curves as special cases. 
For example, 

when k — — 1, a conic with origin at the focus ; 

when k = — 2, a conic with origin at the center ; 
when k — 1, b =fc 0, the limacon ; 
when k = 1, 5 = 0, a circle ; 
when k = 2, b = 0, the lemniscate. 

78. Determination of the orbit when the law of force is known. 

When the law of the force is known as a function of the distance />, 
we may determine the orbit by the integration of equation (4), 
Art. 77. The differential equation is linear and of the second 
order. The general solution will contain two arbitrary constants. 
In general the form of the orbit depends upon the constants of 
integration, which depend upon the initial conditions of the 
motion. The method of integration of the differential equation of 
the orbit depends upon the form of the function F, that is, upon 
the law of force. We shall consider in detail the case of an 
attractive force inversely proportional to the square of the distance. 

In this case F= — = — k 2 mu 2 and the differential equation 

of the orbit becomes ^ 



Kdv 1 J m 



Cm li n 

This is a well-known differential equation of which the solution 
(74, Chap. XIV) is 

h 2 

U=-C 1 COS (0 + C 2 ) + Tg- 

Hence n 



(1) P = 



k 2 



— - c 1 cos (6 + <? 2 ) 1 - c 1 — cos (0 + <? 2 ) 



174 THEORETICAL MECHANICS 

Equation (1) is the equation of a conic section with focus at the 

origin.* The principal axis of the conic makes an angle — c 2 with 

h 2 
the axis of coordinates. The eccentricity is e = a, — • The distance 

.... k A a 

p from the focus to the directrix is given by the relation ep = —, 

whence p = — • 
c i 

Theorem. The orbit of a particle subject to an attractive central 
force varying inversely as the square of the distance is a conic section 
with focus at the center of force. 

The special case of the circular orbit is obtained when we 

h 2 
select c-, = 0. The radius of the circle is — • 

k 2 

To determine the type of the orbit, we must find e in terms of 

the initial distance and the initial speed. Since F = — , the 

P 2 
energy equation (Art. 75) gives 

-K(y 2 - V) = / -— Y d P' 

" v Po r 

Hence 

(2) v 2 = v 2 , 

P Po 

2Jc 2 
from which the result is derived that v 2 — - — has a constant value 

for any particular orbit. ? 

Taking for the equation of the orbit 

_ ep 

1 — e cos# ' 
we find, by differentiation, 

dp _ e 2 p sin d6 _ p 2 sin 6 dd __ _h„- n n 

dt (1 — e cos 0y dt p dt p 

since p z — = h . 
V ^ dt J 



* The standard form of the equation of a conic section with focus at the origin is 

(Analytic Geometry, p. 173) 

p = S£ 

1 — e cos d 

This equation takes the form (1) if the polar axis is rotated through an angle c 2 . 



CENTRAL FORCES 175 



*=m+o' 



h 2 . a/) , h 2 

= — sm 2 + — 

p l p z 



h 2 
e 2 p 2 \ p 



('-* + *-?) 



u 2 2 A; 2 A 2 , 2 ,. , 2 /A 2 72 \ 

Hence v 2 = — — (e 2 — 1)+- A; 2 . 

p e z p* p\ep J 

Since the first member of this equation is constant and inde- 

h 2 

pendent of p, we mnst determine h so that k 2 = 0, that is, 

ep 

h 2 = epic 2 . The above equation now becomes 

(3) v 2_^ == A( e 2_ 1 N 

p ep 
For the three types of conies, (3) gives : 

parabola e = l, .*. v 2 = 

P 

e < 1, a = semi-major axis = e P , 
. ^2 = ^f2_l 



/> a. 
hyperbola e > 1, a = semi-transverse axis, = " , 

^2 = £2(^ + l\ 



From these results we see that at a given distance p from the cen- 
ter of force the speed in an elliptic orbit is less than the speed in a 
parabolic orbit. Also in an hyperbolic orbit the speed is greater 
than in a parabolic orbit. When e = 1, then, by (2) and (3), if 

2Jc 2 
v = 0, then p = oo and v 2 = - — , that is, the speed in a parabolic 

P 

orbit is the speed which would be acquired by a particle starting 

from rest at an infinite distance, or, briefly, the speed from in- 
finity. Hence the 

Theorem. The path of a free particle in an attractive central 
field of force varying inversely as the square of the distance is an 
ellipse, parabola, or hyperbola according as the initial speed is less 
than, equal to, or greater than the speed from infinity. 



176 THEOEETICAL MECHANICS 

79. Position in the orbit. If the orbit of a particle is given 
or has been determined, it remains to determine the position of 
the particle in the orbit at any instant. Let the equation of the 
orbit be p =/(0) . From the law of areas, 

a) ?«-*; 

whence 



(2) ff\0)dd = F(ff) = ht + e s . 



This equation determines the vectorial angle in terms of the 
time. Solving for 0, 

6 = 0(0- 

If the position at any instant is known, the constant of inte- 
gration may be determined and the position at any other instant 
may be found. 

Since is known as a function of t, this value may be sub- 
stituted in the equation of the orbit and p will be expressed in 
terms of t: p — ty(f)* We may now rind the speed at any 
instant from the energy equation, or from the third theorem of 
Art. 73. 

The time T required to describe any given arc from 6 = 1 
to = # 2 may be found from (1) by integration. This gives 



hT=h(t 2 -t 1 )= f e2 p 2 d6, 

(3) ° rT= \ rv m 



This result might have been anticipated from the law of areas. 
The integral in (3) is twice the area bounded by the curve and 
the radii vectores = 1 and 6 = 2 . The constant h is twice the 
areal velocity. Hence (3) may be written 



m area 



areal velocity 

For example, if the orbit is an ellipse, the period, that is the 
time to describe the complete curve, is 

rp _ 2 irah 



CENTEAL FOECES 177 

Illustrative Example. A particle describes a logarithmic spiral under a 
center of force at the pole. Find the time of describing any arc. Determine the 
coordinates and speed in terms of the time. 

Solution. The equation of the curve is p = e ae . From (3), 

hJdi 2ah K J 2 ah K ' J 

From (2), Ce^dd = t + c 3 , 

whence — e 2a& = — p 2 = t + c 3 , 

2a 2a 

and 2a6 = log 2 a(t + c 8 ). 

We may find the speed directly from the relation 

m 2 + P 2 (dey = (dp cwy + 2 (djy = ri /^y + n p4 /<**y. 

Now, -£ = ae a9 — ap, 

dd 

at 



Hence v 2 = h 2 (*±1) = h *\ * + * 1- 

V p 2 ; L2a(« + c 3 )J 



80. Complete solution of a problem in central motion. We 
have seen (Art. 76) that the problem of determining the motion 
of a particle in a central field of force demands the solution of a 
system of two simultaneous differential equations each of the 
second order. The complete solution must contain four con- 
stants of integration. For the determination of the constants we 
must have four initial conditions, for example, the two coordinates 

of position p , O and the two components of velocity (-/) , ( — - ) 

\atJo \dtJo 

at the instant t = t . We must be able to express the constants 

of integration in terms of the initial conditions. 

Griven the law of force, and the initial conditions p = a, = /3 5 

—l = 7, — = S, when t=0,to determine the motion completely. The 

do (XL 

solution of the differential equations of motion (1), Art. 73, is 
accomplished by three steps. 

I. Integrating the second equation, we have 

CD ,§-*. 



178 THEORETICAL MECHANICS 

II. Integrating the differential equation of the orbit, 

we obtain 

(3) P~f&; o v c 2 -), 

which is the polar equation of the orbit and involves two constants 
of integration. 

III. Substituting in (1) the value of p from (3), we have, by 
integration, 

(4) CpaS = F(0)=ht + c 3 . 

To determine the four constants of integration (h, c v <? 2 , <? 3 ), 
we impose the initial conditions. 

I. From (1), 

(5) h = o?8. 

II. To determine c x and c v we differentiate (3) with respect 
to t, 

r „K dp_ df(6^ dd _ ff (Qm dd 

From (3) and (3'), 

We find c x and <? 2 by solving the simultaneous equations (6). 

III. From (4), 

(7) * 3 = ^(/3). 

When the values of the constants of integration given by (5), 
(6), and (7) are substituted in (3) and (4), we have the finite 
equations of motion, by which p and 6 are expressed in terms of t 
and the given constants. 

Illustrative Example. A particle is subject to an attractive central force 
inversely proportional to the square of the distance ( F= — ^ ) . Determine the 

motion completely if p = a, = 0, =£- = 0, — = 6, when t = 0. Discuss the form 

dt dt 

of the orbit for various values of b. 



CENTRAL FORCES 179 



Solution. By (5) the constant of areas is h = a 2 b. 
Hence the differential equation of the orbit is 



W# 2 J 



The polar equation of the orbit is (see (1), Art. 78) 

(8) P = «£ 

v ' 1 - cia*6 2 cos (0 + c 2 ) 

Differentiating (8) with respect to t, we have after simplifying, 

(9) | = _ Clp2sin( , + C2) |. 

Substituting the given initial values in (8) and (9), the equations for the 
determination of C\ and c 2 become 

no) J 1 — Ci«*6 2 cosca' 

[ =— Cia 2 &sinc 2 . 
The solution of (10) gives 

ci= — — , c 2 = 0. 

Substituting these values in (8), the equation of the orbit becomes 

K } P 1- (l-a 3 & 2 )cos0* 

To express 6 in terms of t, we have, from (4), 

where e = 1 — a 3 6 2 . 

Integration of (12) gives 

^L M^-: + -4= arc tan { JT + i tan »1 1 = a% + Ca . 

1 — e 2 Ll — ecosfl Vl — e 2 1*1 — e 2jJ 

Substituting the initial values of 8 and t, we find 

c 3 = 0. 
Hence is expressed in terms of t by the relation * 

^jf e,me 2 arcta ; ( /l±ltan|n=«. 

l_ e 2Ll — e cos 6> VT3^2 (\l-e 2 M 

Equation (11) shows that the orbit is a conic section with focus at the origin. 
For various values of the initial angular velocity &, the following cases occur. 

* The solution of this equation for 6 is not simple. For practical purposes it is cus- 
tomary to employ infinite series. The student is referred to Moulton's Celestial 
Mechanics, Chapter V. 



180 



THEORETICAL MECHANICS 



(1) If & 2 <^, the orbit is an ellipse (4) If b 2 = —, the orbit is a 

parabola. a 



(e = 1 — a B b 2 ) with the left-hand focus at the 
origin. 

CD 



(T~~ 


>V* 


OP = a = l, 62=}, 


n= 


P X 

= a6 = J 



«> 



% 


\t 




V** 






p 







X 



(2) If b 2 = — , the orbit is a circle with Gen- 
es 3 



ter at the origin. ^) 




OP = a =l, 62=1, r = ab = l. 



(3) If i-< 6 2 <— , the orbit is an 
a 3 a 3 

(e = #52 _ i) w i t h t j ie r ight-hand focus at the 
origin. 

(3) 



OP = a = l, &2 =2 , V = ab = lA. 



(5) If 6 2 >— , the orbit is an 

hyperbola (e = a 3 b 2 — 1), with the 
left-hand focus at the origin. 

(5) 




V a =2 



OP = a=l, & 2 = f, V = ab = 1.2. 



OP 



1, 6 2 = 4, V = ab = 2. 



CENTRAL FORCES 181 

PROBLEMS 

1. Determine the various orbits for the law of inverse cube of the distance, 



F _ _ mk 2 

p8 Ans. When k 2 < ft 2 , - = a cos (cd + B). where c 2 = 1 - -. 

P v J h 2 

When k 2 = U\ - = ad + jS. 

1 1.2 

When k 2 > ft 2 , ± = ae^ + be~ ce , where c 2 = — - 1. 
p ft 2 

2. Determine the coordinates and the speed in terms of the time, and the time 
of describing any arc when the orbit is the curve given : 

(a) the reciprocal spiral pd = a ; 
(6) the curve p = ae~ ad 4- be a0 ; 

(c) the lituus p 2 6 = a 2 ; 

(d) the lemniscate p 2 = a 2 cos 2 6 ; 

(e) the cardioid p = a(l + cos 0); 
(/) the limacon p = b — a cos ; 

(#) the four-leaved rose p = a cos 2 ; 
(ft) the three-leaved rose p = a cos 3 6 ; 
(i) the rose p = a cos w0. 

3. When the force isi^=m( — + — ) show that, if v < ft, the general equation 

V/3 2 PV 
of the orbit described has the form 

a 

p = , 

1 — e cos (Jed) 
where a, e, and k are constants. 

4. Show that (if e < 1) the curve in problem 3 may be regarded as an ellipse 
whose major axis rotates about the focus with uniform angular velocity 

27T(l-fc) 
n ~ kf ' 

where T is the time required for the particle to completely describe the rotating 
ellipse. 

5. Show that in the case of a central force the motion along the radius vector 
is defined by the equation 

<Pp _ f _h 2 
dt 2 ' p* 

6. A particle is subject to an attractive central force proportional to the 

distance (F = — mp). Determine the motion completely if p = «, 6 — 0, — = 0,- 

dt 

cld -=b,whent = 0. rtan0 = 6tan*, 

dt Ans. J o a 2 b 2 



b 2 cos 2 d + sin 2 d 



182 THEORETICAL MECHANICS 

7. A particle is subject to an attractive force inversely proportional to the fifth 
power of the distance. Determine the motion completely in the two cases : 

(«) p = aJB ■= 0, ■& = 0, &£ = i when t = 0. 
v J H ' \U clt a s 

(6) p = a, = 0, ^ = 0, <W = _J_. when t _ 
<^ d^ V2 a 3 

-4w5. (a) 6 = t, p = a. 

(6) 2 0, 4- sin 2 = ^|, p = a cos 0. 

81. Planetary motion. The law of gravitation. The astron- 
omer Kepler (1571-1630) was led to formulate the following 
empirical laws of planetary motion, his conclusions resulting from 
the study of a great number of observations made by his prede- 
cessors and himself. 

I. The radius vector of each planet ivith respect 'to the sun as 
origin sweeps over equal areas in equal times. 

II. The orbit of each planet is an ellipse with the sun at a focus. 

III. The square of the period of revolution is proportional to the 
cube of the major semiaxis. 

Upon the basis of Kepler's laws Newton proved that the plan- 
ets move under the action of a force directed towards the sun, 
and varying inversely as the square of the distance, thus. By the 
first law the theorem of areas holds, and we conclude, by Art. 74, 
that the planets are subject to a central field of force with center 
at the sun. There is no evidence that the intensity of the force 
of the field is different for different directions, and we assume 
that the law of force depends only upon the distance. From the 
second law the equation of the orbit is (Analytic Geometry, 
p. 173) ep 

1 — e cos 6 

Therefore, since u = —, 

P 

. d 2 u_ 1 

u + de*~Tp 

and equation (4), Art. 77, for the determination of the law of force, 
gives 

(1) F=-^. 

epp z 



CENTRAL FORCES 183 

We therefore see that, assuming the force depends only on the 
distance, the first two laws of Kepler lead to the conclusion that 
any one planet is attracted by the sun with a force inversely 
proportional to the square of the distance. 

By means of the third law we show that the factor — is the 

J ep 

same for all the planets. By Art. 79, the period T= , 

, 2wab 
or h=-^-, 

( 2 )' and — = TFT' 

ep epT 2, 

TO 

But^> = — (Analytic Geometry, p. 185), and (2) becomes 

ae 

W 4 ttV 



ep T 2 

a s 
Since by the third law — — is constant for all the planets, we 

obtain for the law of force, 

(3) F--e% 

r 

where c has the same value for all the planets. From (3) we may 
conclude, as did Newton, that the sun exerts upon a planet a force 
of attraction which is directly proportional to the mass of the 
planet, and inversely proportional to the square of its distance 
from the sun. 

Laiv of universal gravitation. It is shown by observations 
that laws corresponding to those of Kepler hold for the motion of 
the moon around the earth, and also for the motion of every 
family of satellites in the solar system. It follows, therefore, 
that each satellite is subject to a central force directed towards 
the primary, and varying inversely as the square of the distance. 
It has been shown also in every case in which the motion of a 
comet has been observed that the path is a conic section with the 
sun at a focus, and that the law of areas holds. These bodies, 
therefore, move under the same law of force as the planets. The 
laws of Kepler and the preceding statements concerning satellites 
and comets, although, of immense importance, are only approxi- 



184 THEORETICAL MECHANICS 

mately true. The errors are comparatively small, but easily per- 
ceptible by observation, and readily explained theoretically upon 
the basis of Newton's law of universal gravitation. This is: 
every particle of matter in the universe attracts every other particle 
with a force which acts in a line joining them, and whose intensity 
is directly proportional to the product of their masses, and inversely 
proportional to the squares of the distances apart. 

Observations show that the orbit of the moon about the earth 
is not an exact ellipse. This is due to the fact that its motion is 
influenced by the attractions of the sun and every other member 
of the solar system. The orbit is approximately an ellipse be- 
cause the moon is very near the earth, and the central force 
directed towards the earth is much greater than all the other 
forces acting. The proper computation shows that Newton's law 
accounts for the motions of all the planets and satellites in the 
solar system, and not a single fact is known to dispute its truth. 
By means of it and the appropriate mathematical processes, we 
are able to predict the positions of the planets and satellites many 
years in advance. We therefore consider its truth to 
be established as far as the solar system is concerned. 

Newton's verification that the force which holds 
the moon in its orbit is the same as that which makes 
an apple fall to the ground is historically important. 
To work this out, we need the theorem that the attrac- 
tion of the earth upon an exterior object is the same 
J E as if its mass were concentrated at its center. Next 
we assume that the attraction of the earth for rela- 
tively small masses is the same as if the latter were material par- 
ticles. Consider, therefore, the attraction of the earth (a) upon 
the moon; (5) upon a material particle upon its own surface. 
By (3) these are 

/n XT Mm-, 4 TrVm* ^ Mm 

where M = mass of earth, m x = mass of moon, m = mass of par- 
ticle, r = mean distance of moon, R = radius of earth, a = major 
semiaxis of moon's orbit, T = moon's period, g = acceleration 
due to gravity at earth's surface. 



CENTRAL FORCES 185 

Comparing the values of cM in F t and F 2 , we get from (4) 

Or, 

,rv _4:7T 2 a 2 

This value of g is accordingly the condition that the law of uni- 
versal gravitation shall hold for the influence of the earth upon 
the moon and falling bodies at the earth's surface. The value of 
g computed from (5) is, in the C. G. S. system, 975, which com- 
pares with the observed value of 981 as closely as is to be 
expected under the approximate conditions assumed. 




X 



CHAPTER VIII 

HARMONIC FIELD 

82. Harmonic central field. We begin with the study of free 
motion in a central field due to a center of force attracting directly 
as the distance. That is, if is the center 
of force, then at any point P, 

(1) Force = mk 2 ■ OP, ' 

where k 2 is the absolute intensity of the field, 
that is, the force on unit mass at unit dis- 
tance. 
The axial components of F are 

F x = Fcos (x, P) = - Pcos 6, F y = Psin (x, P) = - Psin 6, 
since (x, P) = rr + 0. But cos 6 = -^-, sin 6 = -^— • Hence 

we have 

(2) F x = — mk 2 x, F y = — mk 2 y. 

Consider now the question of the path of a free particle pro- 
jected with any velocity into such a field. It is obvious that the 
path is a straight line if the initial velocity is along a line of force. 
In the general case, however, the path is curvilinear, and we can 
at least foresee that it must be everywhere concave toivards the center 
of force, since the force causing the motion has that direction. 
The general statement is contained in the 

Theorem. The path of a free particle in a harmonic central 
field is elliptic if it is projected with a velocity oblique to the lines of 
force. 

Proof. The force equations in rectangular coordinates reduce 
by (2) to 

(3) £ + *■-«, g + % = 0. 

186 



HAEMONIC FIELD 



187 



Each equation is harmonic (71, Chap. XIV), and the solutions 
may be written 

(4) x = c x sin Jet + c 2 cos Jet, y — c z sin Jet + c± cos Jet, 

in which c v c, 



2' °3' 




are constants of integration. The rectangu- 
lar equation of the path is obtained from (4) by eliminating t. 
We may, however, simplify the problem if we draw the axis of x 
through the initial position. Then if t = 0, we have y = 0, and 
hence c 4 = 0. We must now eliminate t from 

(5) x = c 1 sin Jet + c 2 cos Jet, y — c s sin Jet. 

This is readily done by solving the second equation for sin Jet and 
substituting in the first. The result is, after reduction, found 
to be 

(6) c 2 x l - 2 c^xy + (C-? + c*)y 2 = c 2 c 2 , 

which is the equation of a central conic with 
center at the origin. But since this conic 
must be everywhere concave towards the cen- 
ter 0, the locus must be an ellipse. Q. e. d. 

Since the rectangular component motions (4) are both periodic 
with the same period 2 ir -^ Je, the particle completely describes 
the ellipse in the time 2 it -j- Je, and we have the important 

Theorem. A free particle projected in a central harmonic field 
in any direction will describe a periodic orbit wJiose period depends 
only upon tJie absolute intensity of tJie field. 

Illustrative Example. An elastic string AB is fastened at A and the other 
extremity is pulled through a ring at B and attached to a heavy particle. The latter 
is at rest in the position C. If the particle is now displaced obliquely a small dis- 
tance and then projected, determine the motion. 

Solution. Let T represent the pull of the string 
when the particle is at D. Then, by Hooke's Law 
(footnote, p. 112), 

(1) T:mg.:BD:BC, 

since at C the pull and weight are equal. The result- 
ant force B is the vector sum of T and mg. The 
vector triangle is, however, similar to BBC by virtue 
of the proportion (1) and the equality of the angle 
at B to that between T and mg. Hence B acts 
towards C and is proportional to DC. That is, the 

particle moves as if attracted towards C with a force proportional to the distance. 

The particle therefore describes a vertical ellipse about C as a center when pro- 




188 THEOKETICAL MECHANICS 

jected from D in the plane BCD in any direction oblique to CD. The period of the 

motion is 2 ir\- ,HBC = d. Tor, since B :mg :: DC : BC, we have B = ^- CD. 
Vg BC 

Hence, if m and CD are unity, the absolute intensity of B is -$— or 2 1 and this 
is & 2 . 

83. Energy equation. The energy equation in rectangular 
coordinates gives, using (2), Art. 82, 



mr — £ mvr 



f ( — mk 2 xdx — m 2 k 2 ydy) 

_ __ mk 2 1~ 
" ~2~L 



# 2 -f- v 2 



x,y w £2 

= - ! ~ (P 2 -Po 2 )> 

J*0»lfo ^ 



if p = distance from center of force. 

(I) .•.V 2 -V0 2 =-^(p2-p^). 

The path (6), Art. 82, is circular, when and only when 
c 1 = 0, c z = c 2 . The equations of motion now are, if c 2 = c 3 = a, 

(1) x = a cos kt, y = a sin H. 

Since p is now constant and equal to p , (I) gives v = v , that 
is, the motion is uniform circular motion. 

84. Simple harmonic motion. The path is clearly rectilinear 
when the direction of the velocity of projection is along a line of 
force. Let s be the distance from the origin at 
any instant. Then F= — mk 2 s, and the force 
equation is 



a) y+ft-o. 



y^o 



that is, the harmonic equation (71, Chap. XIV). 
The solution, or equation of motion, may be written 

(2) s = a cos (kt + /3), 

in which a and ft are arbitrary constants. The characteristics of 
the motion have been discussed in example 3, p. 50. 

The following terminology is in common use for simple har- 
monic motion. 

The attraction F is called the force of restitution, the distance 
s the displacement, and the constant /3, upon which the initial 



HAEMONIC FIELD 189 

position (s = a cos /3) depends, is named the epoch. As already 
pointed out, the maximum displacement (=#) is the amplitude, 
and the period of the motion is 



(3) T = 



2tt 



Since in any given harmonic field the period T is constant, we 

2 7T 

may write in (2), k = ^— , and thus obtain the equation of motion 
in the standard form 

(II) « = acos^ + p)- 

Frequency. The reciprocal of the period is named the fre- 
quency of the vibration. Obviously, the frequency gives the 
number of total vibrations (integral and fractional) performed in 
unit time. 

Phase. The extreme position A is reached when ^— + ft = 0, 



or t = — ■£— T. Let M be a subsequent position in the path at the 

J 

2tt 



2tt - b 

time t. Then the elapsed time from A to, M is £ 4- ■— T. The 



r 



M A 

ratio of this interval to a complete period is called the phase; that 
is, 

(4) Phase at the time t = — + -@- 

For example, if the phase = J, the particle is at since a 
quarter period has elapsed from A ; if the phase = J, the position 
is A f , etc. 

Difference in phase. Given the simple harmonic motions with 
the same period, 

(5) x = a cos f-|p + /3 V a;' = a' sin (-£- + £'\ 

to determine the difference in phase. We must first write the 
second equation in the standard form, thus: 

x> = a' sin P|« + p\ = a, cos &* + &+ 1 



190 THEORETICAL MECHANICS 

The respective phases are now (by (4)) 

£' + - 
_ + JL and - + -g_- 



7T 



/3-/3'-| 

Their difference is accordingly equal to ; 

2 7T 

that is, a constant independent of the time. This gives the result: 
If two simple harmonic motions have the same period, their difference 
in phase is constant. 

Illustrative Example. A heavy particle is at rest at on a rough horizon- 
tal plane midway between two points C and C. The extremities of an elastic string 



£ 



p c' 

of length less than OC are attached to C and to the particle, and a like elastic 
string is attached to C and to the particle. The particle is now displaced a small 
distance from the position in the direction CO and then released ; determine the 
motion. 

Solution. Let d = elongation of each string when the particle is at 0. Then 
at P, if s = OP, the elongations are respectively d + s, d — s. Since the pull of 
each string is proportional to the elongation, the force of restitution, towards 0, is 
numerically m\(d 4- s) — m\(d — s) = 2 m\s, where X is a constant factor of propor- 
tionality. This is resisted, however, by the friction ixmg, where /x is the coefficient 
of friction. The resultant force is the difference. The force of restitution must be 
written — 2 mXs, since its direction is opposite to s. The friction, however, must 
remain indeterminate in sign, ± fimg, since its direction reverses with the change 
in direction of the motion. The force equation therefore gives, after division by m, 

(1) H=_2A S ±^, 

the plus or minus sign being used according as the direction of motion is negative 
or positive. 

Let the particle be displaced from O to A. Then, for motion to begin, the 
force of restitution 2 • Xm • OA must exceed the friction. That is, 2 X • OA > fig, or 



A JJAO D A A 

OA>^fxg -^- X. If then the points D and D' be marked such that OB = OD' = ^2, 
the initial position must be beyond D or D' . 



Separating the two cases in (1), we have for motion in the negative direction, 

^ + 2X S =^,or^| + 2Xf S -^ 

dP dt 2 V 2 



HARMONIC FIELD 191 

If in this equation we write s — ~^- == s ; , we obtain the harmonic equation 

d 2 s' ua 

h2 \s f = 0. We therefore have a harmonic oscillation whose center is s = £&- or 

dt 1 2X 

2) ; that is, on the positive side. 

The solution of the problem is now clear. The effect of friction is to pull the 

center of force in its own direction from to D or D'. Thus, if motion begins at A, 

the particle moves to A', where DA' = DA. The particle next moves to A" such 

that D'A' = D'A'J, etc. We note that friction reduces the amplitude each time by 

DD' or fig -=- a. Motion ceases when an extreme position falls within DD', in the 

figure at A'". 

PROBLEMS 

1. Integrate the following harmonic equations, under the given conditions : 
(a) —-\-x = 0; x = 2,t7o=0. Ans. x = 2 cos t. 

(5) <Ey + 4 y = o ; a = 2, p=Z. Ans. y = 2 cos (2 t + 4 tt). 
d£ 2 6 

(c) 2j? + 3ac = 0; -a=l, 0. = -. ^ws. x = sin V3 t. 

(d) -=5 + w 2 ^ = ; ?/o = c, v = wc. 

dt 2 e 

(In this problem, w = — = angular velocity.) Ans. 6 = a cos -v? £. 

dt * e 

(/) ?? + 2 = O; O = O, u = fc. -4ns. 6 = kJlsh 

dt 2 e Vg 

(g) H+&2 = 0; a = a? ^ = |^ 

2. Show that each of the following defines a simple harmonic motion, by 
writing each in the standard form (II). Find the amplitude, epoch, and period. 

(a) x = sin t — 2 cos t. 1 9 

-4ms. a = V5,sin/3 = ^,cos/3= =^, T=2tt. 

Vo V'5 

(6) x= -cos 2£ + 3 sin 2^. ,4ns. a = VlO, sin ,3 = ^,008/3= — 

VlO VlO' 

(c) »=2siii(* — -Y J.ns. s = 2cos(£-Itt). 

(cH s = cos 7r« — 4 sin 7rt .4ns. a= Vl7, sin (3 = — == , cos/3 — — . 

Vl7 Vl7 



V 



(e) 2 = 5 cos (irt - — Y 

(/) x = cos f tt£ + ^\ - sin f tt£ - |Y 



192 THEORETICAL MECHANICS 

(g) X = d! cos (kt + j8i) + o 2 cos (kt + j8 2 ) . 
(h) y = b\ sin kt + b 2 sin (kt — ir) . 

(i) ?/ = sin £ — cos £ + sin ( t + - ). 

(j ) = a x cos (ni — j8i) + a 2 cos (w* + j8 2 ). 

(A) x = ai COS (#tf + |8i) + o 2 cos (fc£ + j3 2 ) + 3 cos (&t + j8 3 ). 

(0 »= -cos -t-Ssm(-t + -\ 

(to) x — 2 sin \ -Kt — cos (| 7r£ — | ir) . 

(w) x = sin - Tt — 5 cos ( - 7r£ 

3 \3 2 y 

Note the characteristic thing : The functions are sines or cosines and when 
occurring together the coefficient of t is the same. 

3. Show that 

x = «i cos (&£ + /3i) + «2 cos (kt + /3 2 ) + • • • + o» cos (At + /3„) 
defines a harmonic motion. What is the period ? 

4. Draw the distance-time diagram for each solution of problem 1, and 
discuss the figure. 

5. Construct the positions of a particle having simple harmonic motion, if the 
amplitude is 2, and phase equals a, |, 2, 1, 1^, 1^, 2|, 5, |, 1|-. 

6. An elastic string supporting a heavy particle hangs in equilibrium, the 
elongation due to the weight of the particle being equal to d. The particle is now 
depressed below the position of equilibrium through a distance c greater than d, and 
then released. Find the height to which the particle will rise after the string ceases 

to be taut. Ano c 2 — d 2 

x±ns. . 

2d 

7. If the heavy particle in problem 6 be acted upon by an impulse sufficient to 
project it downward from the position of equilibrium with the velocity v, find the 
maximum extension of the string. . , Id 

8. Find the velocity of the impulse in problem 7 if the string just resumes its 
original length when the particle rises. ^ nSt yf^n, 

9. What is the nature of a field of force due to two centers of equal absolute 
intensity and each attracting directly as the distance ? 

Ans. A harmonic central field of double intensity whose center is the middle 
point of those given. 

10. Two material particles act as centers of force attracting as the distance, the 
absolute intensity of each equaling the mass of the particle. Determine the nature 
of the field. 

Ans. A similar field due to the entire mass placed at the center of mass. 

11. Generalize problem 10. 

12. Find the path under a repulsive center of force, the law of direct distance 
still holding as in Art. 82. Ans. Hyperbola. 



HARMONIC FIELD 193 

13. Two heavy particles of masses m and m', respectively, hang at rest, being 
attached to the lower extremity of an elastic thread, whose upper end is fixed. 
Supposing the second particle drops off, determine the subsequent motion of the 
other. 

Arts. If the separate particles cause elongations d and d', respectively, and if 

I = length of string, the equation of motion is x • = I + d + d' cos -J^- t, where 
x = distance of the particle from the fixed point of suspension. 

85. Composition of simple harmonic motions in a given field. 
Many problems in mechanics depend for their solution upon the 
following simple principle. 

Consider two simple harmonic motions occurring simultane- 
ously on XX' in the given field. Their equations may be written 

(1) Zj = a x cos (Jet -f- ySj), x 2 = « 2 cos(&£ + /3 2 ). 

By composition (or addition) we derive from these the motion 
whose equation is 

(2) x = x x H- a? 2 = a x cos (Jet + ft{) + a 2 cos (Jet -f- /3 2 ). 

But this motion is also simple harmonic with the same center 
and period as the components (1). For the equations (1) are 
solutions of the harmonic equation, 

(3) g + fe=0, 

and their sum is also a solution. 

Hence the equation (2) must be in the form 

(4) x= a cos (Jet + ft) . 

In order to find a and ft, expand (2) and (4) and compare the 
coefficients of cos Jet and sin Jet. We obtain 

(5) a cos /3 = a x cos @ ± + a 2 cos fiy a sin ft = a x sin ft 1 + a 2 sin ft 2 . 
Squaring and adding these gives 

(6) a 2 = a^ + a 2 2 + 2 a x a 2 cos (ft 1 — /3 2 ). 

This is the amplitude of the resultant motion. Knowing a, 
then sin ft and cos ft are given by (5) and hence the epoch can be 
found. 

The preceding discussion may be generalized and gives the 
result : 



194 



THEORETICAL MECHANICS 



The resultant of any number of simple harmonic motions on the 
same line in a given field is also a simple harmonic motion in the 
same field. 

Consider next the composition of simple harmonic motions 
in a given field along lines mutually perpendicular. Let these 
equations be 



a> 




x = a cos (Jet + /^i), y — b cos (Jet + /3 2 ). 

The resultant motion is that of the point 
(x, y), and, as already proved, the motion is 
in general elliptic. The path may, however, 
be rectilinear, namely, if the difference in 
phase is J or any multiple thereof; that 



is, if 



(8) 



A -ft . 



n 



, and hence /3 1 = irn + /3 2 , 

2 7T 2 

we get in (7), by substitution, 

(9) x = a cos (Jet + /3 2 + nir~) = ± a cos (Jet + /3 2 ) = ± ^S 

and the path is accordingly a straight line. 

This result is important and gives the 

Theorem. Two simultaneous simple harmonic motions along 
perpendicular lines in the same field compound into a simple har- 
monic motion when the difference in phase is a multiple of one half. 
Conversely, any simple harmonic motion in a given plane field 
may be resolved in two simple harmonic motions along perpendicu- 
lar lines, one of which may be chosen arbitrarily. 

Consider finally the composition of simultaneous simple har- 
monic motions along oblique lines LL' and XX in a given field. 
Let P be any position resulting from 
composition of the motions of M along 
OX and N along OL. By the theorem, 
the motion of N along LL 1 may be re- 
solved into simultaneous simple har- 
monic motions of M' along OX and N ! 
along OY. But since in the figure 
0M n = OM' + OM, the simultaneous simple harmonic motions 
of W and M compound into a simple harmonic motion of M" . 



L' 



Y 


L, 


N' 


K/___^ 




x 1 S '• 


so 


MM M" X 



HARMONIC FIELD 195 

Hence the resultant of the motions of M along OX and N along 
OL is now compounded of simple harmonic motions along the 
perpendicular lines OX and OZ, and is therefore an elliptic 
harmonic motion in general. 

The conclusion of the whole matter may now be stated in the 

Theorem. The resultant of any number of simple harmonic 
motions having the same center and period is either an elliptic 
harmonic motion or a simple harmonic motion. 

This theorem finds numerous applications in Physics in connection with elastic 
media and the theory of wave motion. By Hooke's Law, any particle of such a 
medium, when the stress causing a displacement is removed, performs small oscilla- 
tions under the action of a force of restitution proportional to the displacement from 
a normal position, and therefore executes simple harmonic motion. The com- 
position of such motions is accordingly of importance in studying the effect of simul- 
taneous disturbances in such media. 

PROBLEMS 

1. Find the equation of the resultant of the following simultaneous motions. 

In all cases determine the resultant amplitude, epoch, and the difference of phase. 

(a) Xi = sin t ; x 2 = cos t. Ans. x = V2cos (t — \tt). 

(6) xi = 2 cos*; Xz = sin ( t + — ]■ Ans. x = 2>cost. 

(c) Xi = 2 cos -irt; x 2 = sm(-vt + -]• 
2 \2 6 / 

(cV) Xi=— cos irt ; x 2 = 2 sin (irt + \ tt). Ans. Difference of phase = i. 

(e) y x = sin \t ; y 2 — 2 cos \ t. Ans. Difference of phase = \. 

(/) Vi = - cos (f irt + \tt) ; 2/ 2 = 2sin(f7r£+f tt). 

Ans. Difference of phase = 0. 

(gr) £i = acosA;£; x 2 = a cos! kt + -^- ]• Ans. x = a cos ( kt + - V 

2. Find the equation of motion of the resultant of 

X\ = acoskt, x 2 = a cos (kt + f tt) , x s = a cos (kt -f- |7r). Ans. x = 0. 

3. Find the amplitude and epoch of the motion whose equations are 

x = a cos (kt + j3), V — b cos (kt + j3). 

Ans. Amplitude = Va 2 + b 2 , epoch = /3. 

4. What is the theorem concerning difference of phase when the resultant of 
two simple harmonic motions along perpendicular lines in the same field is a uni- 
form circular motion ? 

Ans. Difference of phase must be an odd multiple of one fourth. 

5. Discuss the motion defined by each of the following. Find the equation of 
the path in each case, and plot the locus. 

(a) x = cost; y = 2sin(£ + \w). Ans. 4 x % — 2 V2 xy + y 2 = 2. 



196 THEOKETICAL MECHANICS 

(6) x = 2 sin t ; y = 3 cos £. Arts. 9 cc 2 + 4 y 2 = 36. 

(c) a; = sin (i 7ri5 + I ir) ; ?/ = — cos (f 7r£). ^4ns. a; 2 + V2 xy + y 2 = \. 

(d) x = cos kt ; 2/ = cos (&£ + f 7r). J.ws. x 2 + xy + y 2 = f . 

6. A heavy particle is suspended from a fixed point by a fine elastic thread 
and is hanging at rest. Motion is set up by an impulse imparting a velocity v in 
a vertical plane through the thread but inclined at an angle a to the latter. Deter- 
mine the motion. 

Ans. Simple harmonic, amplitude = v ■*■ k, where k has same value as before 
(Art. 82). 

7. If the particle in problem 6 is not originally at rest but performing vertical 
vibrations, determine the motion when the same impulse acts upon it when in any 
position. Work out the equations of motion by composition. 

8. Solve problem 3 by rotating the axes through the angle 6 = tan -1 - , and 
show that the new equations of motion are a 



x' = x cos — y sin 6 = Va 2 + b 2 cos (Jet + j8), y r = x sin d + y cos 6 = 0. 

9. A particle is projected from the point (3, 4) with a velocity v of 30 ft. per 
second in the direction given by (v , x) = 1 ir + sin -1 f . The force acting is an 
attraction from the origin varying as the distance and in magnitude equaling 1 lb. 
per unit mass at the distance of 2 ft. Discuss the motion. 

Ans. Path is 9± x 2 - 23f xy + 16 & y 2 = 625. 

10. Discuss the resultant of two simple harmonic motions on the same line in 
a given field if the difference of phase is \ ; \\ \\ f ; f . 

11. If the fly wheel of an engine revolves with constant speed, show that the 
motion of the piston is more nearly simple harmonic the greater the length of the 
connecting rod. 

86. Composition with different periods. Forced vibrations. 

Consider again the motion of a heavy particle performing small 
vertical oscillations by virtue of being suspended by an elastic 
thread. Let the particle be acted upon by a 'periodic vertical 
force, that is, a force whose magnitude and direction vary periodi- 
cally. Such a force is given by 

(1) F = F cos \t = mf Q cos \t. 

Its maximum value is F and its period is 2 ir -*- \. 

It is clear that the original simple vibration will be altered. 
In particular, if the force varies in such a manner that its direction 
always coincides with the direction of motion, the amplitude will 
increase with each oscillation. The periods of the harmonic 
motion and the force must in this case be equal. On the contrary, 
if the periods are nearly equal, the direction of the force will 



HAKMONIC FIELD 197 

eventually oppose the motion and reduce the amplitude. Thus 
a great oscillation will be at first produced, then reduced, and 
subsequently renewed, etc. 

Mathematical verification of these observed facts is readily 
made. The differential equation of motion is 

(2) ^ + ft=/ cosX<, 

since the forces are F and the attraction of the field (= — mk 2 s). 

Case 1. Periods unequal (X =£ F). A particular integral of 

f 
(2) is without difficulty seen to be -~^ — - cos Xt. Hence, the 

rC — A- 

equation of motion is (see 75 (a), Chap. XIV) 

(3) s = a cos (Jet +/3) + -^> cos Xt . 

This equation is obviously obtained by composition of simple 
harmonic motions of different periods, namely, 

(4) s 1 = a cos (Jet + ft), s 2 = ^° cos Xt. 

fC — A, 

The first of these is the undisturbed harmonic motion. The 
second has the same period as the disturbing force (2 7r-;-\), and 
the amplitude is 

(5) b = — £> 

The resultant motion (3) is, of course, an oscillation, but not 
harmonic. The case is important when k and X are nearly equal, 
that is, when the disturbing force F has a period differing slightly 
from the period of the field. The amplitude b is now very large 
and the oscillations of the motion (3) consequently become very 
great. This result may be formulated : 

If a vibrating body is acted upon by a periodic force of frequency 
nearly equal to that of the undisturbed vibrations, the forced oscilla- 
tions will be of great amplitude. 

We have here an illustration of the principle of resonance. 

The conclusion may now be drawn that a small force with the 
proper period may produce remarkable effects, and an explana- 
tion is arrived at of the danger to bridges from the steady march- 
ing of troops, the heavy rolling of ships caused by waves of proper 



198 THEORETICAL MECHANICS 

period, etc. The phenomenon of " beats " in acoustics rests upon 
this principle also. 

Case 2. Periods equal (\=F). A particular solution of 
(2) is now found to be *sl t sin Jet, and the general solution may 

therefore be written in the form (see 75 (5), Chap. XIV) 

(6) s = a cos (Jet + /3) -f A t sin Jet. 

2 Je 

The presence of t in the second term destroys the harmonic 
character of the component. It is plain, however, that this term 
determines the numerical magnitude of s when t is large, and 
consequently it is seen that the amplitude of this vibration be- 
comes and remains very great. 

87. General harmonic field. If the rectangular components 
of a plane field are 

(1) F x = — Jc 2 mx, F y = — l 2 my, 

the field is a general harmonic field. 

The force equations for a free particle are 
in this case 




^ dfi~ ' dt*~ ly ' 

Each of these equations is harmonic, and therefore the equa- 
tions of motion of a free particle in a general harmonic field are 

(3) x = a cos (Jet + /3), y = b cos (It + 7). 

The path evidently lies within and touches the sides of a 
rectangle whose sides are 2 a and 2 b. -Fur- 
thermore, the path will be a closed curve 
when Je and I are commensurable. For if 



Je m r -, . , x 

- = — (m and n integers), 
I n 



N 



M 



X 



then, since & = 2 7r -r- T v 1= 2tt -?- T 2 , where 

T x and T 2 are the periods of the component motions (3), we have 

(4) mT x = nT v 

Hence when a period of time equal to T= mT x = nT 2 has 



HARMONIC FIELD 199 

elapsed, x and y have their original values, and the particle has 
returned to its original position. 

The curves defined by (3) are known in Physics as Lissajous' 
Curves, from the name of the scientist who first studied them. 
They may be defined as the path of a free point whose motion 
is compounded of simple harmonic motions along perpendicular 
lines.* 

PROBLEMS 

1. Obtain the equations of motion of a free particle in the harmonic fields for 
which h = 1, I = V2 ; k = V2, I = 1 ; Jc = VS, I = 1; Jc = 1,1 = 2; k = 2,l = 2. 

2. Determine the path of a free particle in a general harmonic field under the 
following conditions: 

(a) h = 1, I =2, a = 2, b = 1, /3 = y = 0. 

(b) k =i, 1 = 1, a = 1, 6 = 1, £=-, 7 = 0. 

(c) k = 2, l = l,a = b, j3= 7 =-. 

2 



* See General Physics, Hastings and Beach'(Ginn and Company), p. 529. 



CHAPTER IX 

MOTION IN A RESISTING MEDIUM 

88. Law of resistance. In the preceding sections the charac- 
teristics of motion in various fields have been determined without 
reference to any resistance to the motion which might be offered 
by the medium in which motion takes place. The law of resist- 
ance must necessarily be established by experiment. For air, a 
study of this law under given conditions of temperature, pressure, 
etc., has been made by numerous investigators. The following 
table will exhibit results found for rotating projectiles of the stand- 
ard Krupp form, the assumption having been made by the experi- 
menter in each case that tl\e resistance varies as some poiver of the 
speed. The first line gives the speed in meters per second, the 
second line the resistance, a, 5, c, J, e, /, g being constants of pro- 
portionality. 

Speed 50 240 295 3T5 419 550 800 1000 
Resistance av 2 ■ bv z cv 5 dv z ev 2 fv 1 - 7 gv l - bb 

The arrangement of the table is intended to indicate that the 
law of resistance holds for all speeds in the interval under which 
is written the expression for the law. Thus for speeds between 
50 and 240 m. per second, the resistance varies as the square of 
the speed, etc. It will be observed that the resistance involves a 
complicated exponent for very large speeds. In any given case 
the law necessarily depends upon the shape of the moving body, 
the medium (water, air, etc.), and the changing physical conditions 
of the latter (temperature, pressure, etc.). We consider in this 
chapter the effect upon the motion of a resistance assumed to 
follow a given law. 

89. Constant field. Resistance proportional to the square of the 
velocity. We may begin by studying the effect of the presence 
of a resisting medium upon motion in a constant field, and, as an 
example, consider the case of a falling body, assuming the resist- 

200 



MOTION IX A RESISTING- MEDIUM 201 

ance to be proportional to v' 2 . The equation of motion is there- 
fore, if distance is measured downwards, 



d 



a) fr g ~^ 

where jjl is a positive factor of proportionality, called the coefficient 
of resistance. Evidently /jl equals the resistance offered to unit 
mass when moving with unit speed. 
Integrating (1), we have 

2V# Vg—^fiv 
For initial conditions, assume s = 0, v = when t = 0. Hence 
(7 = 0, and solving for v, we obtain, after simple transformations, 

( 2 ) v =\- 1= -J=- ' 

\ /J, /*gt + e v *Qt 

Writing v = — and integrating, we obtain 

Clo 

(3) s = -log ^ 

This is therefore the desired equation of motion. The formula is 
applicable to motion in any constant field under the given condi- 
tions, if g is replaced by the acceleration of that field. It should 
be observed in equation (1) that the initial acceleration under the 
given conditions (v = when t = 0) equals g. The acceleration 

r g 

then diminishes and will be zero if v = \ — Examination of (2), 

however, shows that the speed approaches this value as t increases 

indefinitely. For this reason this value is called the limiting 

speed. In words : the speed increases constantly and approaches 

\~g 
the limiting value \ — 

If the resistance offered by water to the motion of a ship is pro- 
portional to the square of the velocity, then (3) may be applied by 
replacing g by the acceleration due to the propelling force of the 
engines. In the same example, if the engines be stopped when 
the velocity is v , the equation of the further motion of the ship is 

(4) | = _m 



202 THEORETICAL MECHANICS 

since the resistance is the only force acting. Integrating with 
the conditions s = 0, v = v when t = 0, we obtain 

(5) " = — "Vt' • = ilogG*V + l). 

The equations (5) may be applied to the problem in question for 
small values of t, this limitation being necessary on account of 
complications (drifting, etc.) which must soon arise. 

90. Damped harmonic motion. Resistance varying as velocity. 

The motion of a material particle in a central harmonic field has 
already been discussed. We now investigate the effect of the 
presence of a resisting medium in such a field. For small speeds 
the resistance may be assumed proportional to the first power of 
the speed. Further, since the resistance and the velocity have 
opposite directions, we may set 

Resistance = — 2 fimv, 

where ft is a positive constant, called the damping factor. The 
force due to the field being equal to — mk 2 s, the resultant force F 
acting upon the particle is 

F= — 2 fxmv — 



fF d 2 s\ 
and the force equation ( — = — - ) consequently may be written in 

\Tfl Ctu J 

the form 

Two important cases present themselves for discussion. 

(a) Damping factor small, fi < k. Equation (1) is now the 
equation of damped vibration (73, Chap. XIV), the equation of 
motion being 

s = Ae-v* cos (V& 2 -/* 2 £ + /3), 

in which A and {3 are arbitrary constants. The characteristics of 
this motion have been discussed at length in example 5, p. 51. The 

2 7T 

motion is therefore a damped vibration with the period — • 

V& 2 -/* 2 

Obviously the effect of the damping factor is to increase the 

2 7T 

period, since the latter for the undamped vibration equals — — • 



MOTION IN A RESISTING MEDIUM 203 

A close approximation to damped vibration is afforded by the 
simple pendulum for small vibrations. To obtain the desired 
result we may use the moment equation 
(Art. 62). Taking for center of moments 
the center of suspension, the total force- 
moment is ' / 

Rl — mgl sin 0, 
where 



R = - 2 m/iv = - 2 mfil— • mg 




R 



The angular momentum = I Q a> = ml 2 — • Hence the moment 

ctz 

equation gives 

— 2 mid 2 - mgl sin = — f ml 2 - — ) = ml 2 — - : 

p dt y dt\ dt) dt 2 

that is, 

d?6 . dO , q . a n 

For small amplitudes, sin 6 = 6, approximately, and this equation 
now agrees with (1). The discussion confirms the observed facts 
of the constancy in the period of a simple pendulum and the de- 
crease in the amplitude. 

(5) Damping factor large, fi>k. The solution of (1) is now 
(Calculus, p. 437) 

(2) s = Ae r S + Be r J, 

in which r x and r 2 are the roots of the characteristic equation 

r 2 + 2 fir + k 2 = 0. 

Let us discuss (2) for the initial conditions s = a, v = when 
t = 0. These conditions are nearly met if a large vertical damp- 
ing vane be affixed to a magnetic needle and if the latter is then 
slightly turned from its position of equilibrium and released. 

Differentiating (2), we obtain 

(3) v = r x Ae r J + r 2 Be r *. 

We have now for t = 0, 

a=A + B, 0=r 1 A + r 2 B, 
and hence 



204 THEORETICAL MECHANICS 

and (2) and (3) become 

(4) g = —^— (r t e r * - r^S), v = ^^ («'V - #f\. 
r x -r 2 r x - r 2 

From these equations the following characteristics of the 
motion are obvious — results agreeing with experience. The dis- 
tance * diminishes and approaches zero as a limit. The speed 
increases to a maximum and then diminishes to zero. 

PROBLEMS 

1. A particle is projected with velocity v Q into a medium offering a resistance 
proportional to the velocity ( = kv) . Show that the particle would come to rest 

after describing the finite space — in an infinite time. 

2. If the resistance of a medium is kv 2 , show that a particle projected with a 
velocity vo would describe an infinite space in an infinite time before coming to rest. 

3. If the resistance of the medium per unit mass is kv 2 , and a particle slides 
under the action of gravity on a smooth straight wire inclined at an angle a to the 
horizontal, prove that the space s described in time t from rest is given by 

/ p2bt i 1 ' 

ks = U-\og{ e -— ± y 
where & 2 = kg sin a. * 

4. A heavy particle is projected upwards with a velocity L in a medium resist- 
ing as the nth power of the velocity. Prove that the whole space (up and down) 
described when the velocity downwards is V is equal to XT where L is the limiting 
velocity and T is the time in which the particle falling from rest in the medium will 

, . V 2 

acquire a velocity 

L 

* Since r 1 (=— fi+Vp. 2 — k 2 ) and r 2 (= — /x — V/u 2 — k 2 ) are both negative, e r i* and 
e r 2 ' both approach zero as t increases indefinitely. 



CHAPTER X 

POTENTIAL AND POTENTIAL ENERGY 

91. A constant, harmonic, or general central field has the 
property : There exists a function U of the coordinates x and y 
such that the rectangular components of the force of the field are 
the negative partial derivatives of this function. That is, 

The function Z7is called the potential of the field. Obviously 
the potential is given as the integral 



(2) U=-J(F x dx + F y dy), 



Constant field. Then F x and F y are constants, say F x = J., 
F y — B\ hence U= — (Ax + By), and conversely, equations (1) 
hold. 

Harmonic field. ■ Here F x = — mk 2 x, F y = — mk 2 y ; hence 
U— — \mk 2 (x 2 + ?/ 2 ), and conversely, equations (1) hold. 

Let it be understood, then, that the potential of a field (if the 
field have a potential) is a function of the coordinates satisfying 
equations (1). 

92. Conservative field. For a field in which there exists a 
potential the following theorem is characteristic: 

The work done by the force of the field upon a material particle 
moving from one position to another is the same for all paths between 
those positions. 

Proof. If a particle moves ^^ Cj ""— — -y$ x &*) 

from P O , y ) to F 1 (x v y x ) along "** — gL ^ 

a path C 1 in the field whose rec- 
tangular components are F x and F y , the work done is (Art. 62) 

f (F x dx + F y dy), 

205 




206 THEORETICAL MECHANICS 

the integral being worked out for the given path. If, however, 
the field has a potential, that is, if equation (2) of the preceding 
section holds, then 

(1) T" V \F x dx + F y dy-) = - ( U x - Z7 ), 

Z7 X and U being the potential at P 1 (x v y-[) and P (^ , # ), 

respectively. But this equation shows that the work done equals 

the difference of the potential at P 1 and at P taken negatively. 

Hence the work done along any other path C 2 is the same, and 

the theorem is proved. 

Again, let the particle describe any closed path from P (# , ^ ). 

The total work done is now zero. 

F(x y )^^ yPi^i'Vi) For take a second point 

PiQc v y{) on the path and de- 
note the path from P to P x 

by O v and from P t to P by 2 . Then by the result just found 

Work done along C 1 = — (U 1 — Z7 ) ; 
Work done along C 2 = — ( Z7 — U-^). 

Adding gives the work done along the closed path as zero. 
The designation conservative is applied to a field possessing a 
potential, and also to the force of such a field. The discussion 
may be summarized thus : 

The work done by a conservative force along any path equals the 
negative difference of the potential at its extremities. 

93. Potential energy. Conservation of energy. Comparison of 
the energy equation (Art. 62) with the theorem just stated gives 
the result : 

When a material particle describes any path in a conservative field, 
the change in kinetic energy equals the change in potential taken 
negatively. 

In the form of an equation, this statement reads 

(1) | m V- 2 imV=-(*7i-tfo)> 

if v 1 and v are the speeds at (x v y^) and (# , ?/ ), respectively. 

By transposing in (1), remarking that \mv^ is a constant, and 
dropping the subscript l, we obtain 

(2) \mv l + ( U- Z7 ) = constant. 



POTENTIAL AND POTENTIAL ENERGY 207 

Now ^mv 2 gives the kinetic energy of the particle at any 
instant. Hence, since each term in this equation must be of the 
dimensions of energy, we may give to (27— Z7 ) the name of poten- 
tial energy, that is, we define 

(3) Potential Energy = TJ - U ; 

or in words : The potential energy at any point in a conservative 
field equals the change in the value of the potential from an arbitrarily 
chosen point of reference. 

The essential difference between kinetic and potential energy 
is this : Kinetic energy is due to motion — depends upon mass and 
velocity. Potential energy is due to relative position — depends 
upon the position relative to an assumed point of reference. 

In (2), writing 

E k = \mv\ U p = U- U , 

we obtain the equation of energy for a conservative field: 

(I) Ek + E p = constant. 

Equation (I) illustrates the Principle of the Conserva- 
tion of Energy for a material particle in a conservative field, 
namely, this : 

If a material particle describes any path in a conservative field, 
the sum of the kinetic and potential energy remains constant. 

A simple illustration of a non-conservative field is afforded by the following 
example. Using polar coordinates, let 

The work integral is now (Art. 62) ° v"* "^^ I 

Work = ( i m 9 Hd = ^ f P *dd ; \ •« f / 

that is, the work done now equals the mass times the \ ; / 

area swept over by the radius vector of the curve in \ I / 

moving from OPq to OPi, a number obviously dependent \y 

on the path c between the points. The field in ques- O 

tion is a simple one, the force at any point P being perpendicular to the radius 

vector OP, and proportional to it. 

94. Equipotential lines and lines of force. If U(x, y) is the 
potential function for a certain field, then at every point of the 
locus of 

(1) U(x, y) — constant 

the potential is the same. 




208 THEORETICAL MECHANICS 

By assigning to the constant in (1) different values, we derive 
a series of curves called equipotential lines, such that the potential 
is the same at all points on one of these curves. 

For example, in the constant field for which U = Ax + By, the equipotential 
lines consist of the system of parallel lines Ax + By = constant. 

If the equipotential lines are drawn in an}^ 
field, the work done along any path joining 
two points P and P equals the difference 
of the potential of the equipotential lines 
through P and P , or equals (c x — c 4 ) in the 

figure. The slope of the equipotential line (1) at any point (x, y) 

is (Calculus, p. 202), 

( 2 ) 2=-|=-j : (b y( ivA r ,9i): 

By 

Let us now find the direction of the force F of the field at the 
point P(x, y). Since the axial components of F are F x and F y , 
then the 

(2') slope of the force F = |f. 

Comparing with (2), it is clear (Analytic Geometry, p. 36) 
that the direction of F is perpendicular to the equipotential line 
through the point of application. 

The system of curves drawn in a field of force such that the 
direction of the curve through any point is the same as the 
direction of the force of the field at that point are called lines of 
force. Clearly, the differential equation of these lines is 

( 2 ") t- = f?< or *Jy - F y d * = °- 

ax M x 

If this equation can be integrated, the lines of force may be 
constructed. Or, if the equipotential lines have been drawn, we 
may construct the lines of force by drawing the orthogonal trajec- 
tories. For, as is clear from the above discussion, we have the 

Theorem. Equipotential lines and lines of force intersect 
everywhere at right angles. 

The coordinates x and y of any point P on a curve are func- 



POTENTIAL AND POTENTIAL ENEEGY 209 

tions of the arc s( = PqP) measured from an assumed initial 
point P . Hence, in a conservative field, the potential U along a 
curve may be considered a function of the length of arc. Then, 
since now TJ along the path PJP is a function of s, we have 
(Calculus, p. 199) 

(3) dU = d_Udx + dUdy 

ds dx ds dy ds 
This becomes, by substitution from (1), Art. 91, 

W ds V x ds^ y ds. 

Remembering that — and ~f- are the direction cosines of the 
ds ds 

tangent to the curve C at P, we see (Art. 40) that the second 

member of (4) gives the tangential component along Q of the 

force due to the field taken negatively. That is, from (4), 

(5) f— J, 

The derivative of the potential ivith respect to the are of a curve 
equals the tangential component along that curve of the force due to 
the field, with sign changed. 

For example, the components of the force of the field parallel 
and perpendicular to the radius vector of the point (/?, 0) are 
found thus : 

Taking the path along the radius vector 
(0 = constant), we have 

ds = dp, .-. ^L=-F p . 
dp 

Taking the path as the circle p = constant, then ds = pdd, and 
hence 

ldU = F 
p dd * 

In particular, if we consider the variation of the potential 
along a line of force, then, in (5), F t is the force of the field, or 
F itself. 

(6) .-. along a line of force, —— = —F. 

as 

In this equation we may assume the direction of increasing 
arc to agree with the direction of the line of force. Then, by (6), 





210 THEORETICAL MECHANICS 

— — is always negative (or zero) and hence TJ is a decreasing 
as 

function.* That is, the potential along a line of force diminishes 
in the direction of the force. Otherwise 
expressed thus : The force at any point in a 
conservative field is directed towards the region 
of lower potential. If, therefore, the equi- 
potential lines are drawn in a field, the direc- 
tion of the force of the field at each point is 
uniquely determined. 
Finally, the equipotential lines (1), when drawn for equal 

increments of the potential, for example, c, c±e, c ± 2 e, c ± 3 e, 

etc., will, if the increment e is small enough, indicate by their 

degree of proximity the relative magnitude 

of the force of the field. For, by the theorem 

of mean value, f we have, using (6), 

(7) AU=-(F} alP -As, 

where F is the force of the field at some 

point P between successive equipotential 

lines, and As is the distance apart of these lines measured along 

the line of force AB. Since A £7 = e, (7) becomes, by solving, 

Hence the force at P is inversely proportional to the normal 
distance (=As) between consecutive equipotential lines. 
Our results are summarized as follows: 

Theorem. When the equipotential lines are drawn in a con- 
servative field for equal small increments of the potential, the force of 
the field at any point is inversely proportional to the normal distance 
between consecutive lines. 

That is, the force is greatest where the equipotential lines are 
most dense, etc. In a field for which the force is everywhere of 

* The function of s increases or decreases with s according as its derivative with 
respect to s is positive or negative (Calculus, p. 116). 

t This theorem maybe stated: Given a function U(s), for which ^L=—F, then 

ds 
U(s) — U(so) =— (F(s'))(s — So), where s' is a value of s between s and s, and F(s') is 
the value of F when s = s'. 




POTENTIAL AND POTENTIAL ENERGY 



211 



the same magnitude, the equipotential lines drawn for equal 
increments are equidistant. 

We now illustrate the preceding by examples. 



ILLUSTRATIVE EXAMPLES 

1. Discuss the conservative field for which the potential is U= a 2 x. 
The equipotential lines are a 2 x = c, that is, 
lines parallel to YY'. Since 



F x = - 



dU 



dU 
dy 



= 0, 



the lines of force are parallel to XX 1 and are 
directed to the left. Setting c = 0, ± e, ±2e, 
etc., we obtain the equidistant equipotential 



lines of the figure, x = 0, x = ± 



x = ± 



2e 



etc. 



The work done by the field along any path from 
3 e 3 e 6 e 




P to T equals 
field. 



a negative number, since the motion is against the 



2. Discuss the conservative field for which the potential is TJ = \ k 2 x 2 . 

The equipotential lines are \ k'-x 2 = c, lines parallel to 
YY'. We find 





Y 














































X 



Y 




X 



F x = 



dU 
dx 



= - k 2 x, F y = 0, 



so that the force at any point is directed towards YY', and 
is proportional to the distance of that point from YY' 
Setting c = 0, ± e, ± 2 e, ± 3 €, etc., we obtain the equipo- 
tential lines of the figure. It is observed that these lines 
are closer together as we recede from YY', an indication of 
increasing force. The potential is a minimum when x — 0, 



and the necessary condition for this, namely, 
x — 0, is seen to be satisfied. 



dU. 
dx 



when 



tial is U 



3. Discuss the conservative field for which the poten- 
k 2 



k 2 



The equipotential lines are the concentric circles — = c, 

P 



or p = k 2 -4- c. "We find F p = — — 



dU 



d P 



- l -,Fe 



P 69 




In the figure the equipotential lines are drawn for c = k 2 , 
k 2 ± e, k 2 ±2e, etc. The decreasing distance between the 
circles as the center is approached indicates increasing force. 

4. Discuss the Principle of the Conservation of Energy for the motion of a 
projectile when the angle of elevation is \ ir. 



212 THEOEETICAL MECHANICS 

Choose the initial position as the point of reference for determining potential 
energy. Then in the equation of energy (I) we have, since at 
E p = 0, E k = I mv 2 , the result 
(1) E k + E P = \ mv 2 . 

The lines of force being directed downwards, E p increases during ascent, 
and diminishes during descent. Hence during ascent E k must constantly 
decrease ; that is, the velocity must decrease, and become zero, namely, at 
the highest point. The maximum value of E p is therefore \ mv 2 . In de- 
scent, since E p constantly diminishes, E k increases without limit, that is, 
the speed increases. 

Analytically, we have for the field 

F x = 0, F y = - mg. 

.'. E p = U- Uo = — y ' Fydy = mgy. 
Hence the equation of energy is 

I mv 2 + mgy = \ mvo 2 . 
If h is the greatest height, then 

mgh = i mv 2 , or h = v 2 -h- 2 g. 

5. Discuss the equation of energy for a simple pendulum. 

Taking the lowest point for reference, then in (I), if v is the speed at O, we 
have for the equation of energy 

(1) E k + E p =imv 2 . 

Since the lines of force are directed downwards, E p in- 
creases in ascent and diminishes during descent. Hence, 
as in the preceding example, E k must diminish when the 
particle moves from towards A ; that is, the speed must 
decrease to zero at A. At this extreme position, E p is 
a maximum. In descent from A, E p diminishes towards 
zero ; E k increases, reaching the maximum value ( = I mv 2 ) again at 0. This cycle 
is repeated from to A 1 and A' to 0. The motion is therefore a ceaseless vibra- 
tion from A to A'. 

Analytically, as before, E p = mgy, and (1) is ^ mv 2 + mgy = | mvo 2 . The 
greatest vertical height is therefore v 2 -h- 2 g. 

6. Discuss the equation of energy for simple harmonic motion. 

Assuming the center as point of reference and v as velocity at 0, the equa- 
tion of energy is 

"A^ ^ T~ (1) E k + E p = ± mvo 2 . 

From example 2, we know that E p increases when the motion is away from 
the center, and diminishes for motion towards the center. Hence E k must con- 
stantly diminish and become zero when the particle moves away from the center, 
and must subsequently increase until the center is again reached. We see, there- 
fore, as in example 5, that the motion must be a ceaseless vibration. 

Analytically, since F x = — mk 2 x, F y = 0, we have 

E p =- C F x dx = I mk 2 x 2 - 

Hence (1) is \ mv 2 + | mk 2 x 2 = \ mvo 2 , and for the amplitude we find 

x = ± Vq h- k. 




POTENTIAL AND POTENTIAL ENEEGY 213 

PROBLEM 

1. Discuss the conservative fields for which the potential function U is that 
given : 

(a) by ; (6) ax + by ; (c) ax 2 ; (d) 6y 2 ; (e) ax 2 + 6?/ 2 ; (/) cxy ; 
(gr) ax 2 + by 2 + 2 cfa + 2 ey. 

95. Non-conservative forces. Friction. The work done by the 
force in a conservative field changes sign when the direction of the 
motion is reversed. It is therefore obvions that friction is not a 
conservative force. For the direction of the force of friction is 
reversed when the direction of motion changes, and conse- 
quently the work done by a frictional resistance does not change 
sign. For example, if the resistance offered by the air is con- 
sidered in the motion of a particle projected vertically upwards, 
the frictional resistance is opposed to the motion in all positions, 
and consequently the work done is negative in both ascent and 
descent. In the illustrative example 4 of Art. 94, therefore, we 
see that in nature the velocity of the projectile will be lessened, 
a conclusion agreeing with the observed fact that the velocity is 
less when the projectile returns to the initial position. Again, in 
the discussion of the simple pendulum (example 5, Art. 94), if 
the frictional resistances present in nature are considered, the 
motion will not be perpetual. Friction will act to diminish the 
velocity. 

In general, non-conservative forces are said to be dissipative, 
since by their action kinetic energy is lessened without an equiva- 
lent increase of potential energy. In nature, non-conservative 
forces are always present, and accordingly the principle of 
the conservation of energy does not apply in the form enunciated 
in Art. 93. To cover the actual facts, thermal and chemical 
energy must be considered, matters with which we are not 
concerned in this volume. 

96. Newtonian potential. According to Newton's Law of 
Universal Gravitation (Art. 81), two particles attract each other 
with a force varying directly as the mass of each and inversely as 
the square of their distance apart. The force is therefore, with 
proper notation and units, 

' ( i) F= _ «*»'. 

p 2 




214 THEORETICAL MECHANICS 

Consider, now, the plane field of force due to the attraction 
of a particle of mass m situated at the origin. The force upon 
unit mass at any point is, therefore, 

(2) F P =-~, F 9 = 0. 
P 2 

The potential function is, accordingly, 

(3) U=C- Cp p dp=C--. 

If, for p = go, we assume U = 0, then (7= 0, and (3) becomes 

(4) Z7=--. 

P 

The newtonian potential at P is defined as equal to m -f- /o, 
or, denoting this by JV, 

(5) N=-U=-' 

P 

From the result of Art. 93, we may make the definition : 
The newtonian potential at any point equals the work done in moving 
up to that point from infinite distance. 

To study the field due to two attract- 0*~ / 

ing centers of masses m and m\ we merely / p ' 

have to observe that the newtonian po- O' 

tential at any point P must equal the sum of the potentials due 
to the separate masses. This appears at once from the above 
definition. Hence, 

(6) at**™ A 

p p' 

Similarly for any number of centers. The newtonian potential 
due to the attraction of a continuous solid is readily defined, for if 
the solid is divided into elements of mass, and a point chosen in 
each element, then, proceeding as usual, we define the potential 
N at P as 



(7) N= T— , 



in which p denotes the distance from P to any point of the solid. 

Illustrative Example. Find the potential due to an attracting thin homo- 
geneous spherical shell of density r. Find the force of the field at any point. 



POTENTIAL AND POTENTIAL ENERGY 215 

Solution. Take the center of the sphere as origin and draw OX through the 
point P. Let OP = c. We may consider the shell divided into strips by planes 
passed through it perpendicularly to OX The mass d m of one of th ese strips * is 
2 iradx times r. The distance p from the strip to P is Vy' 2 + (c — x) 2 . Hence 

dx 



r a 2ira-rdx n C a 

N ■= I — = 2 irar I - 

J -a vV 2 + (C — X) 2 J -a 




Va 2 + c z — 2 ex 

The radical must be taken with the positive sign, and this makes it necessary to 
distinguish two cases. 

(I) c~>a, exterior point. Then 

Ar 27r«r r , . x-, 4ira 2 T mass of shell 

JN = \c — a — (c + a)\ = 

c c 

Hence the newtonian potential for a spherical 

shell at an exterior point is the same as if its 

mass were concentrated at its center. 

(II) c< a, interior point. Now 

that is, is the same at every interior point. Hence 
the force of the field is zero within the shell. 

The result for an exterior point is extended at 
once to a solid sphere by conceiving it to be made up of concentric shells. Hence 
the 

Theorem. The newtonian potential for a solid sphere at an external point is 
the same as if its mass were concentrated at its center. 

Consider now an interior point of a solid sphere, at the distance of x from its 
center. We conceive this sphere as consisting of (a) a spherical shell of interior 
radius x and exterior radius equal to that of the solid sphere ; (6) a solid sphere of 
radius x. 

Take these up in order. 

(a) Since the potential is everywhere constant within the shell, its attraction 
is zero. Consequently, the attraction exerted is that due to the solid sphere of 
radius x. 

(6) The mass of this sphere is f 7rrx 3 , and hence the attraction is by the theorem 
— f 7ttx 3 -h x 2 , or equal to — f tttx. That is, the attraction exerted by a solid sphere 
upon a point within it is proportional to the distance of that point from the center 
of the sphere. 

Finally, we readily see that the mutual force of attraction of two solid spheres 
is the same as if their masses were concentrated at their respective centers. 

PROBLEMS 

1. Find the newtonian potential due to a line distribution of matter (thin 
uniform rod) at a point on the line produced. 

Ans. N = rlogf 1 + - ) , where I = length of line, d = distance of point from 
nearest end. * ' 

* The surface of a zone equals the product of its altitude by the circumference of a 
great circle. 



216 THEORETICAL MECHANICS 

2. Find the force of attraction of the rod in the first example upon unit mass 
at the given point. A vr _ dN _ r 

dx d(l + d) 

3. Find the newtonian potential due to a homogenous circular disk at a point 
on the line through the center of the disk and perpendicular to its plane. 

Ans. i\T= — — ( y/ a i _|_ x 2 _ x ^ w here m — mass of disk, x = distance of point from disk. 

4. Find the force of attraction exerted by the disk in problem 3 upon unit mass 
at the given point. . -,_ 2m f x __ A 

a \y/ a 2 + x 2 / 

5. Find the newtonian potential due to the attraction exerted by a homogeneous 
right cylinder or cone at a point upon the axis. 

6. Find the newtonian potential due to a homogeneous square plate at a point 
on a side of the square produced. 



CHAPTER XI 

SYSTEM OF MATERIAL PARTICLES 

97. System in a plane. The preceding chapters have been 
devoted to the study of motion of a single particle. We shall 
now study the simultaneous motion of two or more particles, and 
in this manner prepare ourselves for the study of motion of a 
solid. 

Consider a system of two particles moving in a plane. 
Wj and m 2 be their masses, and the positions 
at any instant P t (x v y^), P 2 (x 2 , y 2 ), respectively. 
Then if (F u , F ly ) are the axial components of the 
complete resultant of the forces acting upon the i?] 
particle at P v and (F 2x , F 2y ) the same at P 2 , 
the force equations (Art. 62) are, for the first particle, 

C 1 ) m i C W = F ^ m i^ = F ^ 

and for the second particle, 

(V\ m ^2 _ XT djh _ XT 




Motion of the center of gravity. Adding the first equations in 
(1) and (2), we obtain 

d jc d oc 

(3) %-^i + m,-^ = F u + F w ; 

also from the second equations we get 

(4) m^ + mf^ = F ly + F 2y . 

If P(x, y) is the center of gravity of the given particles, then 
(Art. 22) 

- = w^a?! + Wgg^ - = m 1 y 1 + m 2 y 2 
m l + m 2 m 1 -f- m 2 

or also 

(5) ra^ + m 2 x 2 = (wj + m 2 >, m 1 y 1 + ra 2 y 2 = (ra x + m^y. 

217 



218 THEORETICAL MECHANICS 

Differentiating (5) twice with respect to t, and substituting 
in the first members of (3) and (4), we obtain 

d x d^u 

(6) (m 1 + m 2 ) w = F lx + F 2x , (m 1 + m^) -^ = F ly + F 2r 

The second member of the first of equations (6) is the sum 
of the JT-components of the resultant forces acting on the first and 
second particles which comprise the system. It is therefore the 
sum of the JT-components of all forces acting upon the system, 
and will be denoted by F x . Similarly the second member of the 
second of equations (6) is the sum of the I^-components of all 
forces acting upon the system, and will be denoted by F y . 
Denoting the mass of the system, which is the sum of the 
masses of the individual particles, by M, equations (6) may be 
written 

(7) M§-f m M§ = F r 

These are the fundamental equations of motion of the center 
of gravity of the system. Their discussion shows that this point 
moves as if forces equal and parallel to the given forces were act- 
ing at that point upon a mass equal to the combined mass. 

That is, the center of gravity moves as if the total mass of the 
system were concentrated at that 'point and acted upon by forces equal 
and parallel to the given forces. 

98. System in space. Let the masses of n particles moving 
in- space of three dimensions be denoted by m v m 2 , ••• m n . Let 
the positions at any instant be given by P x (x v y v 2 X ), P 2 (x 2 , Hv Z( ^-> 
••• P n (x n , y n , s ra ), and the axial components of the complete resultant 
of the forces acting upon the individual particles by (F lx , F ly , F l2 ), 
(F^, F 2y , F 2Z ), ... (F nx , F ny , F nz ). The force equations for the 
separate particles are 

d 2 x 2 _ „ d y 2 _ d z 2 _ p 

(1) I *~oW~ ■**» 2 dt* " 2y ' 2 d& " *' 



, ^L -V w —lb - F m ^ 

' n dt 2 ""' n dP " ** n dt 2 



SYSTEM OF MATERIAL PARTICLES 219 

Motion of the center of gravity. Adding the first column of 
equations (1), we get 

( 2 ) m l~^t + W 2-Jr + - m n-^r = F u + F 2X + .- F nx . 

If P(x, y, z) is the center of gravity of the system, then 



n~n 



m 1 + m 2 -f- ■••»*„ 

If M(^— m 1 -{- m 2 + ••• w n ) is the total mass of the system, and 
-Fp(= J 7 ^ + .F 2;r + ■■'F nx ) is the sum of the .^-components of all 
forces acting on the system, it is evident that equation (2) may 
be written 

*§-*' 

Similar equations in y and z respectively follow from the 
second and third columns of (1). Hence the fundamental equa- 
tions for the motion of the center of gravity of the system are 

(4) Jff-J' - ,Jfg-J' i ,lff-Ji. 

Equations (4) show that the statement of the preceding arti- 
cle concerning two particles moving in a plane holds also for n 
particles moving in space. Hence the 

Theorem. The center of gravity of a system of material par- 
ticles moves as if the total mass of the system were concentrated at 
that point, and acted upon by forces equal and parallel to the given 
forces. t, 

In particular, if the vector sum of all ^1 

forces acting is zero, then y* T 

d?x_ n dty^r. d*z_ () fi^ 

dp' ' d?" ' &■' ^> 



xl 



and the center of gravity moves with uniform 
rectilinear motion. For example, in the case ** 

of three particles moving under the action of their mutual gravi- 
tational attraction, the forces are in pairs equal in magnitude but 
opposite in direction, and their vector sum is zero. Hence the 



220 THEORETICAL MECHANICS 

Theorem. The center of gravity of a system of particles mov- 
ing under the action of their mutual gravitational attraction will de- 
scribe a straight line with constant speed. 

This result is called the principle of the conservation of the 
motion of the center of gravity. 

99. Moment equation for a system of particles. Let the 

first of equations (1), Art. 97, be multiplied by y x and the sec- 
ond by x v Then, by subtraction, 

CD ^^-y^y^-y^. 

It may be easily verified (see Art. 59) that 

d 2 y l _ cPx 1 _ d f dy^ _ dx x 



Xl d$ Vl dP dt 



( X d V\ _ y dX l\ 



Hence equation (1) may be written (see Art. 59) 

d f dy-, dx, 

x.m-, -^i — y-,m-, —J 

dt\ x x dt &1 1 dt 



(2) » ( x^m - yi m^l ) = x,F ly - Vl F v 



Now the second member is the moment (Art. 58) of F x with 
respect to the origin, and the quantity in parenthesis is the 
moment of momentum (Art. 59) of m x with respect to the origin. 
Denoting the latter quantity by H v equation (2) becomes 

—=-^ = moment of F-,. 

dt l 

Also from equations (2), Art. 97, similarly, 

-^~ = moment of F . 

dt l 

If H = H x + H 2 denotes the total moment of momentum of the 
system, and if the total force-moment of the system is defined as 
the sum of the moments of the resultant forces acting on each 
particle, we have, by addition, 

dTT 

-=- = total force-moment. 
dt 

This result is clearly a generalization of (VIII), Art. 59.* 

* This result, proved for two particles in the XY-plane, may be extended by a similar 
process to any number of particles in space of three dimensions. 



SYSTEM OF MATERIAL PARTICLES 221 

In particular, if the forces acting are as before mutual gravita- 
tional attractions only, the forces and consequently their moments 

cancel in pairs, and jjj- 

— — = 0, or 
dt 

H — constant. 

Theorem. Principle of the Conservation of Moment 
of Momentum. The total moment of momentum in any system 
of particles moving under the action of their mutual gravitation is 
invariable. 

100. Work and energy of the system. The total work done 
upon a system of material particles is obtained by summing up 
the work done upon each individual particle by the resultant of 
the forces acting upon it. The kinetic energy of the system at 
any instant is the sum of the kinetic energy of the individual 
particles. Referring to the two particles of Art. 97, we have for 
any displacement, 

Work done by F t = ^(w^ 2 — m 1 v^ 2 ') = change in K.E. of m v 

Work done by F 2 = \{m^ 2 — m 2 v 2 ' 2 ) = change in K.E. of m v 

Hence, by addition, 

(1) Total work done = \\_m\V^ + m 2 t>2 2 - (miVi' 2 + m 2 V2' 2 )] 

= change in K. E. of system. 

As in the preceding articles the result expressed by equation 
(1) is general. 

101. Rigid system of particles. An especially important ex- 
ample of motion of a system of particles arises when the system 
is rigid; that is, when the mutual distance 
of each pair of points is invariable. The im- 
portance of this case is due primarily to its 
application in the case of a rigid solid body, 
which is then regarded as a continuous rigid 
system. 

The rigidity is to be regarded as maintained by a constraint 
which exerts upon any pair of particles P 1 and P 2 reactions equal 
in magnitude but opposite in direction. These reactions are un- 
known, but cancel out in the equations of motion, as shall now 
appear. 




222 THEOEETICAL MECHANICS 

The forces acting in the motion of any rigid system may be 
classified as 

(1) reactions due to the rigidity ; 

(2) impressed forces. 

Let QR lx , R ly ) be the axial components of the reaction due to 
rigidity at P v and (i? 2i27 , R 2tJ ) De ^ ne components of this force at 
P 2 . Then 

(1) R lx + R 2x = ; R ly + R 2y = 0. 

Let the sum of the axial components of the impressed forces at 
P x be (F lar) F ly ) and at P 2 be (P 2x , F 2y ). Then the equations of 
motion are, for the first particle, 

and for the second particle, 

d oc cL • \i 

( 3 ) ™2^| = ^W+^ m 2-^= F 2y + R 2 y 

Then the method of Art. 97 gives at once the 

Theorem. The center of gravity of any rigid system moves as 
if the mass of the entire, system were concentrated at that point and 
acted upon by forces equal and parallel to all impressed forces. 

In analytic form this theorem is for space 

(I) M% = F , M%= Fn M§ = F „ 

where M is the total mass of the system, F is the resultant of the 
impressed forces, acting at the center of gravity (a?, y, z). 

For a rigid system of particles in the X]T-plane we may obtain 
by the process of Art. 99 the 

Theorem. The time-rate of change of the total moment of mo- 
mentum of any rigid system equals the total force-moment of the im- 
pressed forces. 

In analytic form, this is 

i=l i=\ 

In applying this theorem, any point whatever may be chosen as the 
center of moments. 



SYSTEM OF MATERIAL PARTICLES 



223 




Consider next the question of work and energy, supposing the 
two particles of Art. 97 are rigidly connected. Let the equations 
of motion of P 1 and P 2 be, respectively, 

KJ yi = fl(0> ^2=^2(0' 

At the instant £=0 let the position of the 
system be P X P 2 in the figure. At any 
other instant, £, let the position be P^P 2 ' . 
The total work done by the reactions 
due to rigidity is then given by the sum of the definite integrals 

/ (R u dx x + R ll/ dy 1 ) + / (R 2x dx 2 + R^dy^ . 

By virtue of equations (4) the integrands are functions of t 
alone, and since from (1), R lx = — R 2x , R ly = — R 2y , we may write 
the expression for the work in the form 

(5) / [R lx (dz 1 - dx 2 ) + R ly (dy 1 - dy 2 y\ . 

Now Jo 

(6) (*i-z 2 ) 2 +G/i-*/2) 2 = ^ 

where I is constant. Differentiating (6), we obtain 

(7) (x x - a: 2 ) (dx 1 - dx 2 } + (y x - # 2 ) (dy x - dy 2 ~) = 0. 
Y 

P2 



Now 



i£ u = R 1 cos (z, = Rjb—Zi 




R ly = R x sin (x, = R i y*—2i. 



x i — 7?" i*' 



~Vi 



R^' 



Hence (7) becomes after dividing out 



iV 



R 1 Jidx 1 - dx 2 ) + R ly (dy 1 - dy 2 ) = 0. 

Hence the definite integral in (5) is zero, and the reactions R 1 
and R 2 do no work. We see therefore that in a rigid system the 
work done is contributed by the impressed forces only. This re- 
sult can be extended at once to a system of n particles, and we 
have 



224 



THEORETICAL MECHANICS 



(III) 
Work done by impressed forces = change in kinetic energy 

n 

= X (nnvi 2 - niiVio 2 ). 

Work done in a constant field upon a rigid system. Let the 
constant force be weight, and draw the axis of Y vertically down- 
wards. Then for a single particle the work done is 

m iff G/i - Y i)i 

if y r and Y t are the final and initial ordinates. Hence the total 
work is 



(8) 



X ^gQVi - Yi) = g X m &* - X MiYi 





If Y and y are the initial and final 
K ordinates of the center of gravity, 
then ^a Tr ^-\ 

Y= v — ' y= v — 

Hence (8) becomes, if X m i — M, 
(IV) Total work done by gravity = Mg(y - Y). 

This gives the important 

Theorem. If a rigid system is in motion under the action of 
weight only, the total work done equals the total weight of the system 
multiplied by the vertical displacement of the center of gravity. 

PROBLEMS 

1. Two particles of masses 50 lb. and 40 lb. are acted upon at a certain 
instant by parallel forces of 75 poundals and 60 poundals, respectively, whose lines of 
action are 4 ft. apart and perpendicular to the line joining the particles. Determine 
(a) the position of the center of gravity and (b) its acceleration at the instant 
named. 

Ans. (&) 1.5 ft. per second per second, if the forces have the same direction. 

2. If the two particles of problem 1 attract each other with forces of 40 
poundals, the remaining data being as before, compute (a) the acceleration of each 
particle and (6) the acceleration of the center of gravity. 

Ans. (b) 1.5 ft. per second per second. 

3. A particle of mass m slides down a smooth inclined plane of angle a, the 
plane itself (mass M ) being free to slide on a smooth table. Find the acceleration 
of the particle and of the plane. 



SYSTEM OF MATERIAL PARTICLES 225 

4. A system consists of two particles, of which one (mi) moves always on the 
X-axis with an acceleration — k 2 x, and the other (m 2 ) along the F-axis with an 
acceleration — k 2 y. Discuss the motion of the center of gravity. 

5. To the system of problem 4 is added a third particle (m 3 ) which moves along 
a line whose inclination to the X-axis is 45° with a constant acceleration a. Discuss 
the motion of the center of gravity. 

6. Three particles in the XF-plane are acted upon by forces as follows : 

mi by a force equal to kt whose inclination to X-axis is 45°, 
ma by a force equal to kt whose inclination to X-axis is 135°, 
ms by a force equal to — y/2kt in the direction of Y-axis. 
Show that the center of gravity moves uniformly in a straight line, 



CHAPTER XII 



DYNAMICS OF A RIGID BODY 



KINEMATICS 

102. Rigid body. A rigid body is defined mathematically as 
a continuous system of material particles whose mutual distances 
remain unchanged. The motion of a rigid body is known if the 
motion of each point of the body is known. More explicitly, we 
say the motion of a rigid body is completely determined if we 
know : 

(1) the position of the body at any instant ; 

(2) the velocity of each point at any instant ; 

(3) the acceleration of each point at any instant. 

The position, velocity, and acceleration of each point are known, 
if the position, velocity, and acceleration * of three of the points 
not on the same straight line are known. Hence in the general 
case, the discussion of the motion of a rigid body may be reduced 
to the discussion of the motion of a system of three particles, 
forming an invariable triangle. For practical purposes we con- 
fine our attention to the simple types of motion treated below. 

103. Translation. The motion of a body is a translation if 
every line in it remains parallel to its original 
position. Such a motion is observed in the 
driving rod of a locomotive or in the motion 
of a book sliding upon a table so that one 
edge of the book remains parallel to one edge 
of the table. At any instant every point of 
the body has the same velocity both in direction 
and magnitude. The motion is completely 

determined if the motion of a single point is known, e.g. the motion 
of the center of gravity. 

* Since the mutual distances of these points are invariable, these quantities are not 
independent. 

226 




DYNAMICS OF A RIGID BODY 227 

In uniform translation the velocity is constant and the path of 
any point is a straight line. 

In uniformly accelerated translation the acceleration is con- 
stant and the path of any point is rectilinear. 

104. Rotation. The motion is a rotation if the body turns 
around a fixed axis, its points describing circles which lie in 
planes perpendicular to the axis and have their centers on the 
axis. An example of rotation is furnished by a fly wheel. At 
any instant, every point of the body has the same angular velocity 
about the axis. The motion is completely determined by the 
motion of a single point. 

In uniform rotation the angular velocity is always constant. 
' In uniformly accelerated rotation the angular acceleration is 
always constant. 

105. Uniplanar motion. In this type of motion the body 
moves so that all its points move parallel to a fixed plane. Ex- 
amples are furnished by a rolling cylinder and the connecting 
rod of an engine. 

The velocity of each point is parallel to this fixed plane, which 
is called the directing plane. Each line in the body perpendicular 
to the directing plane moves parallel to itself, and at any instant 
every point of this line has the 
same velocity. Consequently, 
we need to study the motion only 
of all points in the body lying 
in a plane parallel to the direct- 
ing plane. Let the plane of the 
paper be such a plane. Let 
and P be any two points. 

Let v = velocity of ; 
v = velocity of P. 

Since the distance OP is invariable, the motion of P relative to 
must be a rotation about 0. If v r is the velocity of P relative 
to 0, then v r must be perpendicular to OP. The actual velocity 
(v) of P is then compounded of the velocity (v ) of and the 
velocity (v r ) of P relative to . That is in the sense of vector 
addition v = v + v r . 




228 



THEOKETICAL MECHANICS 



The velocity of any other point P' in the section may be 
treated in the same way. The velocity (v' r ) of P' relative to 
must be perpendicular to OP' and, since the body is rigid, we 
have the proportion v r : OP : : v' r : OP', The actual velocity (V) 
of P r is given by v'= v' r + v . Hence the motion of the body is 
at any instant compounded of: 

(a) a rotation about a temporary axis chosen arbitrarily per- 
pendicular to the directing plane ; and 

(b) a translation parallel to the directing plane with a velocity 
equal to that of any point on the temporary axis. 

In the notation used, 

Velocity of translation = v ; 



(i) 



Angular velocity of rotation a> = 



OP 

Theorem. Instantaneous Axis. There is at each instant 
an axis perpendicular to the directing plane which is at rest. 

Proof. Draw AB in the plane of the 
section at right angles to v . Assume the 
direction of rotation of AB about as in 
the figure. Lay off on AB 




(2) 



00 = \ 

(O 



where v = speed of 0, a = angular ve- 
locity about 0. Then the velocity of rota- 
tion of C about = — v . Since the actual 
velocity of C is the resultant of v and — v , 
it is zero. q.e.d. 

The point is the instantaneous cen- 
ter of the section. The locus of the in- 
stantaneous centers is the instantaneous 
axis. Since the velocity of (any point) 
is v = co • 00 by (2), the angular velocity 
of the body about the instantaneous axis is co also, that is: 

Theorem. In the resolution of a uniplanar motion into a trans- 
lation and a rotation, the angular velocity about the axis is the same 
for all axes. 

To construct the instantaneous center. Given the actual veloci- 




DYNAMICS OF A RIGID BODY 



229 



A 






ties of two points and 0' in a section parallel to the directing 
plane. Let OT and O'R be the vectors representing the veloci- 
ties. From the preceding proof of the exist- p , 
ence of an instantaneous center C, it is seen 
that must lie on a line perpendicular to OT, 
and also on a line perpendicular to O'R. Draw 
OL and O'U perpendicular to OT and OR, re- 
spectively. Then the intersection of OL and 
O'L' is the instantaneous center. To construct 
the instantaneous center in any uniplanar 
motion it is necessary to know only the directions of the motion 
of two of the points. q.e.f. 

In the preceding discussion we have proved the 

Theokem. Uniplanar motion may at any instant be regarded 
as a rotation about the instantaneous axis. 

106. Centrodes. The instantaneous center moves both relative 
to the body and in space. The locus of its various positions 

relative to the body is called the body 
centrode SS. The locus of its va- 
rious positions in space is called the 
space centrode S'S r . The body cen- 
trode is fixed relative to the body. The 
space centrode is fixed in space. 
These two loci are tangent at any 
instant. The motion of the body 
may be arrived at by rolling the body centrode upon the space 
centrode. 




107. Screw motion. In this type the motion is compounded 
of a translation and rotation. The body rotates about an axis in 
space and at the same time undergoes a translation along the axis. 
The motion of the points in the axis of rotation is clearly a trans- 
lation. The path of any other point is a curve traced on a 
cylinder about the axis of rotation. This curve is a helix if the 
angular velocity bears a constant ratio to the velocity of trans- 
lation. The position of the body at any instant is given by the 
position of one of its points (provided this does not lie on the 
axis of rotation). 



230 



THEORETICAL MECHANICS 



ILLUSTRATIVE EXAMPLES 

1. The motion of a projectile is compounded of a uniform translation along its 
axis and a uniform rotation around its axis. Find the equations of motion of any 
point. 

Solution. Let the Z-axis be the line of motion of the axis of the projectile. 
Consider the motion of the point P(x, y, z). A plane section through P per- 
pendicular to the Z-axis has for its instantaneous center the point (7(0, 0, z). The 
motion of G is uniform translation. Hence, 

z = a+ bt. 

The motion of P relative to C is uniform rotation. 
Hence, if d denote the distance CP and if the initial 
position of P is in the XZ-plane, we have 

f x = d cos kt, 
[y = d sin kt. 

Hence the equations of motion of the point P are 

r x = d cos kt, 
y = d sin Jet, 
z = a + bt. 

The path of the point is a helix (Calculus, p. 272). 




2. A line AB moves with its extremities on two perpendicular lines. Find the 
centrodes and the direction of motion of any point at any instant. 

Solution. Let the point A move along the X-axis and the point B along the 
F-axis. The instantaneous center corresponding to any position of the line AB is 
found by erecting a perpendicular to the X-axis at A and a perpendicular to the 
Y-axis at B. These lines intersect in the point G, 
which is the instantaneous center. The body cen- Y 
trode is the locus of the point C relative to AB. B 1 
Since G is the vertex of a right triangle constructed _g 
on AB as a hypotenuse, the body centrode is a 
semicircle (AGB) with AB as diameter. The space 
centrode is the locus of G in space, that is, rela- 
tive to the XF-plane. Denoting the coordinates 
of G by (aj, y) and the length of AB by I, we have, 
for any position of G, 



x 2 + y 2 = I 2 . 




A 



A' X 



Hence the space centrode is a circle (A'CB'), with center at the intersection of 
the two fixed lines and radius equal to the length of AB. It is readily seen that the 
motion of AB under the conditions stated in the problem may be accomplished by 
rolling the circle ACB, of which AB is the diameter, on the inside of the circle 
A'CB', of which AB is the length of the radius. 

At any instant the motion of the line is a rotation about G. Hence the direc- 
tion of motion of any point P is perpendicular to the line PG. 



DYNAMICS OF A RIGID BODY 231 



PROBLEMS 

1. The center of a fly wheel moves in a straight line (in the plane of the wheel) 
with constant velocity &, while the wheel turns with constant angular velocity w. 
Find the equations of motion of a point on the circumference. 

2. In the preceding problem suppose the center moves with constant accelera- 
tion /, and the wheel turns with constant angular acceleration a. Find the equa- 
tions of motion of a point on the circumference. 

3. A circular disk rolls on the X-axis. If the center moves with a constant 
acceleration/, find the equations of motion of a point on the circumference. 

4. A circular disk A rolls on the exterior of a second circular disk B. Deter- 
mine the centrodes. 

5. A pole slides through a fixed ring while one end moves along a horizontal 
line a feet below the ring. Determine the centrodes. 

6. A chord of a circle moves around the circumference. What are the 
centrodes ? 

7. Construct the centrodes for the connecting rod of an engine. 

8. A point P of a plane figure moves with constant speed along a straight line 
while the figure rotates with constant angular velocity. Show that the body and 
space centrodes are respectively a circle whose center is P and a line parallel to the 
path of P. 

9. A coin of radius a rolls down a plane. What is the locus at any instant 
of all points having the same speed as the center. 

10. A plane figure moves' in its own plane so that a point P moves on a curve C 
with constant speed, the figure meanwhile rotating with constant angular velocity. 
Show that the motion may be obtained by rolling a circle with P as a center upon 
a, parallel curve of C. 

11. Construct the centrodes for problems 8 and 10; (a) when the acceleration 
of P along its path is constant and the angular velocity is constant : (6) when the 
acceleration of P is constant and the angular acceleration is constant. 

KINETICS 

108. Force equations. Work and energy. In Art. 101 cer- 
tain theorems on the motion of a rigid system of material particles 
were proved when the number of particles is finite. These theo- 
rems can be extended to cover the motion of a rigid body, which 
has been defined as a continuous rigid system of material particles. 
In the case of a finite number of particles the theorems were 
proved (see, for example, Art. 97) by forming the sum of a finite 
number of expressions. In the case of an infinite number of 
particles forming a continuous system, the limit of the sum is con- 



232 THEORETICAL MECHANICS 

siderecl. In other words, the ordinary finite sum is replaced by 
the definite integral. This process, which is carried out in detail 
in the following article on kinetic energy, shows that the theo- 
rems of Art. 101 are applicable to the motion of a rigid body. 
For the motion of the center of gravity we have the 

Theorem. When a rigid body is subjected to the action of any 
forces, its center of gravity moves as if the entire mass of the body 
were concentrated at the center of gravity and the given forces applied 
there parallel to their former directions. 

For example, when no forces are acting, the center of gravity 
has uniform motion in a straight line. When the forces acting 
are all equal and parallel, the center of mass has uniformly accel- 
erated rectilinear motion, or else describes a parabola. The center 
of mass of a projectile describes a parabolic orbit. 

From the theorem stated above we may write the force equa- 
tions for the motion of the center of mass. For plane motion of 
the center of gravity these are : 

where M is the total mass of the body, (x, y) the coordinates of 
the center of mass, and F is the resultant of all applied forces. 
From Art. 101 we have for a rigid body the energy equation 

^ TT x Work done on a rig-id body 1 ~, . , . , . 

(11) _ „. ® * \ = Chang-e in kinetic energy. 

v J by all impressed forces J _ &,T 

In particular 

Work done by weight = Mgh, 

where M is the mass of the body and h is the vertical distance 

described by the center of gravity. 

109. Kinetic energy. The kinetic energy of a system of 
material particles was denned as the sum of the kinetic energy 

of each particle, __ _ j» . 

r K.E. = V|mrf. 

»=i 
We now apply this definition to the continuous system of particles 
forming a rigid body. Dividing the whole mass in any way into 
n small elements A 2 m, A 2 m, •••A l m, ••• A n m, we have as an approxi- 
mate value of the kinetic energy of the small element A t m the 
expression \A{mv^ 



DYNAMICS OF A RIGID BODY 



233 



where v { is the speed of some point P { within the element. As an 
approximate value of the kinetic energy of the whole mass we 
have n 



A„ = 



V J Aimv? 




Let the number of elements into which the 

whole mass is divided be increased indefinitely 

in such a way that A t -m (for every %) approaches 

zero as a limit. Then the definition of the kinetic energy 

the rigid body is 

limit , limit ^a 1 



of 



(HI) K.E. = 



n 



i=i 



where the definite integral is understood to extend over the entire 
mass. 

We shall consider the calculation of the kinetic energy in the 
four types of motion treated in Arts. 103-107. 

(i) Translation. For every point of the body the velocity is 
the same at any instant. Hence in (III) v is a constant, and 

(1) K.E. = 1 Cv 2 dm = v l Cdm = i Mv\ 

where Mis the total mass of the body. 

(ii) Hotation. For every point of the body the angular velocity 
co about the axis of rotation is the same at 
any instant. Consider an element of mass dm 
at P, moving with velocity v in the circle 
whose center is 0. Then if co = angular ve- 
locity about the axis ?, 




rco. 



Hence 

(2) K. E. = I fv 2 dm = j co 2 Cr 2 dm 

where I t = moment of inertia with respect to I. 



ii*?, 



(iii) Uniplanar motion. Through any point choose a tempo- 
rary axis ?, perpendicular to the directing plane. It was shown in 
Art. 105 that the motion is compounded of (1) a motion of trans- 
lation with velocity v equal to the velocity of 0, and (2) a motion 
of rotation with angular velocity co (the same for all points) 



234 THEORETICAL MECHANICS 

about L The kinetic energy due to translation is J Mv 2 and the 
kinetic energy due to rotation is J Itf^- Hence the total kinetic 
energy is 

(3) K.E. = i*/ + |V. 

If the temporary axis is the instantaneous axis, then v = and 

(4) K.Tl. = iW, . 

where I c is the moment of inertia about the instantaneous axis. 

An important formula for the kinetic energy is obtained if the 
temporary axis passes through the center of gravity of the body. 

Then (3) becomes 

(5) K.E. = i¥V+lV- 

This formula exhibits the kinetic energy as made up of the 
energy of translation of the entire body with a velocity equal to 
that of the center of gravity and the energy of rotation about an 
axis through the center of gravity. 

(iv) Screw motion. Since screw motion is compounded of a 
translation along an axis in space and a rotation about this axis, 
the total kinetic energy is the sum of the energy of translation and 
the energy of rotation. If v denotes the velocity of a point in the 
axis, the kinetic energy of translation is J Mv 2 . If I is the moment 
of inertia with respect to the axis along which the body is moving, 
and a) is the angular velocity about this axis, the kinetic energy 
of rotation is Jift> 2 . Hence the total kinetic energy of a body 
executing screw motion is given by 

K.E. = |i)^ 2 + iift>2. 

110. Moment equation in rotation. It was shown in Art. 99 
for any system of material particles that the time derivative of the 

moment of momentum with respect to a 
point is equal to the total force-moment 
with respect to that point. This result 
may be stated in a new form when the 
system of particles is rigid and rotates 
about an axis. Consider first a single 
particle in the JTY"-plane moving in a 
circle of radius r about the origin. Then 
the moment of momentum, Art. 59, with respect to the origin is 




DYNAMICS OF A RIGID BODY 235 

mrv where v is the linear velocity, or mr^co where co is the angular 
velocity. Since m and r are constants, the moment equation of 
the particle is 

(1) mr z —- = mir—- n — I, 

v J dt dt 2 

where I is the moment with respect to of the resultant of all 
forces acting upon P. Now mr 2 is the moment of inertia, i, of m 
with respect to 0. Hence (1) may be written 

« «£-«. 

Equation (2) holds for each particle of a rigid system. Suppose, 
for simplicity, the system consists of two particles, P 1 and P 2 , 
rigidly connected, and rotating about the origin. Since the system 

is rigid, the angular acceleration — - is the same for each particle. 



Hence we have 



and by addition, 



dt 2 

.d 2 6_ 1 .d 2 0_ 7 
h dt 2 ~ lv h dt 2 ~ 2 ' 



d 2 e 



( 3 ) (*l + *2>^=Zl + * 



V 



Since the moments of the internal forces cancel in pairs, the second 
member of equation (3) contains only the sum of the moments 
of the impressed forces, which is denoted by L. Further i^i^ — I 
is the moment of inertia of the system. Hence we have the 
moment equation in rotation, 

(IV) xf|=x. 

By the process of integration employed in deriving (HI), it is 
readily shown that (IV) holds when the system of particles forms 
a rigid body. Hence the 

Theorem. The product of the angular acceleration and the 
moment of inertia of a rigid body with respect to an axis about which 
it is rotating is equal to the sum of the moments of the impressed 
forces with respect to that axis. 



236 THEORETICAL MECHANICS 

111. Comparison of formulas in translation and rotation. 

Rotation 
Translation 



Uniform Translation. 



v = v c 




Uniform Rotation. 



CO = COr 



8 = 8 o + V- 




e = #o + "V- 


Uniformly Accelerated, 




Uniformly Accelerated. 


v = v o +A 




co = co + cc£, 


s=s + v t + ±ft 2 . 




0=<9 O + o> £+!<*£ 2 . 


f = acceleration. 




a = angular acceleration. 


K.K. = \Mv\ 




k.e. = i.z;a> 2 . 


Dree Equation 


Moment Equation 


d 2 s xt 

m— := F. 

dtf 




<ft 2 


Hence, to 
Translation 


change formulas in 

to Rotation 


Replace linear velocity- 


by 


angular velocity. 


Replace linear acceleration 


by 


angular acceleration. 


Replace mass 


by 


moment of inertia. 


Replace distance 


by 


angle. 


Replace force 


by 


moment of force. 



112. Fundamental equations for uniplanar motion. It has been 
shown (Art. 105) that uniplanar motion may be regarded as com- 
pounded of a translation with the velocity of an arbitrarily chosen 
point 0, and a rotation about an axis through perpendicular to 
the directing plane. If for the point we chose the center of 
gravity, the motion of translation is determined by the force 
equations (I). The motion of rotation is determined by the 
moment equation (IV), and since the angular velocity is the same 
for all axes, the choice of the axis is arbitrary. Hence the funda- 
mental equations for uniplanar motion are 



DYNAMICS OF A RIGID BODY 



237 



(V) 



(It 2 



(Force Equations; 
(Moment Equation) 



Work done by all impressed forces 

= change in kinetic energy. 

(Energy Equation) 



K.E. = hMv c 



■Igto 2 > 



In equations (V) the point (#, y) is the center of gravity of the 
body, v g is the velocity of the center of gravity, I g is the moment 
of inertia with respect to the gravity axis perpendicular to the 
directing plane, L is the resultant moment of all the impressed 
forces with respect to any axis perpendicular to the directing 
plane, and i"is the moment of inertia with respect to the same 
axis. 



ILLUSTRATIVE EXAMPLES 

1. Compound Pendulum. A heavy body is suspended on a horizontal axis 
and swings under the action of weight. Determine the motion. 

Solution. Let G' he the extreme position of the center of gravity, Go he 
the lowest position of the center of gravity, and G be any position of the center of 
gravity. Consider the motion from G' to G. 

From (2), Art. 109, 

K.E. = \ W, 

where Ia is moment of inertia about the axis of sus- 
pension, and w is the angular velocity when the center 
of mass is at G. 

The work done by weight when the center of 
mass falls from G' to G is 

Work done = mg • MN. 
But MN = AN - AM = AG cos d - A G' cos a. 
Let AG = AG' = d, 

.'. Work = mg • d(cos 6 — cos a). 
By the energy equation, 

mg • d (cos 6 — cos a) = | Ia^ 2 - 
. 2mg-d , 

Ia 
dd 

2 mq • d , 




Hence w 2 - 

Differentiating with respect to t, since o = 



(cos 6 — cos a), 



'dt 2 



0J = 



V dl 



238 THEORETICAL MECHANICS 

df) 
After division by w = — , this equation becomes 
dt 

dt 2 l A 
This agrees with (6), Art. 68, if 

md 
Hence the 

Theorem. A compound pendulum moves precisely like a simple pendulum 

whose length is given by the formula I = — ■ 

md 

The corresponding simple pendulum is called the equivalent simple pendulum. 

2. A homogeneous circular cylinder of mass M and radius r, rotating about its 
axis a times per second, falls from rest through a vertical distance of h feet under 
the action of its weight. Compute the total kinetic energy. 

Solution. The cylinder is executing uniplanar motion and hence the kinetic 
energy is given by (V) . The moment of inertia of the cylinder with respect to its 
axis is I g = ^ Mr 2 . The angular distance moved in one second is 2 -rra radians. 
Hence w = 2 air. The kinetic energy of translation is found at once from the 
energy equation to be 

\ Mv 2 = Mgh. 
Applying (V), 

K.E. = £ • I Mr 2 (2 air) 2 + Mgh. 

.-. K.E. = M (Wtt 2 + gh). 

In many problems involving a system of two or more connected 
bodies acted upon by given external forces the motion may be 
discussed by the previous methods if we take into account the 
reactions due to the connections. From the fundamental equa- 
tions thus obtained for each body the reactions or internal forces 
may be eliminated and the motion determined. The process is 
illustrated in the following example. 

3. A cord passes over a smooth peg as shown in the figure. To one end of the 
cord is attached a mass m, which falls vertically, and the other end is fastened to the 
axle of a solid disk of mass M and radius B which rolls on a plane inclined at an 

angle a to the horizon. Discuss the motion. 

Solution. The motion of the system is known 
if the acceleration of m is known. If T denotes 
the pull of m on the string, then the resultant 
force acting on m is mg — T. The forces acting 
on the disk are (1) the pull of the string T, 
(2) the weight Mg, (3) the friction F at the 
point of contact with the plane, (4) the normal 
pressure N (not indicated in the figure) of the plane on the disk. The pressure 
N acts normal to the plane at its point of contact with the disk. We suppose that 
the friction is sufficient to prevent slipping. Then its point of application is the 




DYNAMICS OF A RIGID BODY 239 

instantaneous center. Since at any instant this point is at rest, the work done by 
the friction is zero. The work done by the normal pressure N is also zero. 

Supposing that to starts from rest, moves through a distance s, and acquires the 
velocity v, the energy equation gives 

(1) I mv 2 = (mg - T)s. 

Applying the energy equation to M, we get (since v is the velocity of the cen- 
ter of gravity of M ) 

(2) i Mv 2 + i Igbfl = (- Mg sin a + T)s. 

Now I g = l MB 2 , and w = -• 

B 
Hence i"X = ± Mv 2 . 

By substitution of this value (2) becomes 

(3) f Mv 2 = (- Mg sin a + T)s. 
Adding (1) and (3) to eliminate the tension T, we have 

(4) (i to + f M)v 2 = (m — M sin a)gs. 

In deriving equation (4) we added (in the second member) the work done 
by all the forces acting on m and the work done by all the forces acting on M. 
The work done by T and — T cancels ; that is, for the system under consideration the 
total work done by the internal forces is zero. Hence for this system we have the 

Theorem. The change in the total kinetic energy of the system is equal to 
the work done by the external or impressed forces . 

Differentiating (4) with respect to t, 

(5) (in + lM)v — = (m-M sin a)g-> 

dt dt 

Since v = — , and — =/, equation (5) gives the acceleration of m, namely, 
ctt dt 

(6) f— (m — Msma)g 

J TO + | Jf 

To find the tension T we differentiate (1) with respect to t, cancel v = — , and 

dv dt 

substitute the value of / = — from (6). The result is 
dt 

V TO + | M J 

The magnitude of the friction al force which is necessary to prevent slipping 
may be found from the moment equation. Taking moments with respect to the 
axis of the cylinder, the moments of the forces Mg, N, and T are zero, and the 
moment equation becomes 

(7) I^=-F.B. 

dt 2 

Since 0=±, then &i = ±^ = ±f. 

B dt 2 Bdt 2 B J 

Also I=\MB 2 . 

Substituting in (7), we obtain 

w- _ i Mf- - ( m ~ J*" sin «) ilfflr 

* /_ 2m+SM 



240 THEOKETICAL MECHANICS 



PROBLEMS 

1. A uniform cylindrical shot weighing 200 lb. is fired from a rifled gun with 
a velocity of 1000 ft. per second. Find the total kinetic energy at the muzzle if it 
rotates 25 times per second, the diameter of the shot being 6 in. What must be 
the average pressure during the discharge if the length of the gun is 7 ft. ? 

Ans. K.E. = [- 2 2 5 -(25tt)2 + los] foot-poundals. 
Pressure = \ (K.E.) poundals. 

2. A uniform circular disk rotates about an axis through its center perpendicu- 
lar to its plane. The disk weighs 16 T. and its radius is 3 ft. (a) What is the 
kinetic energy when it is revolving at the rate of 200 revolutions per minute ? 
(&) What constant tangential force must be applied to a crank 18 in. long to give 
the disk this speed from rest in 1 min. ? (c) If the disk lifts a weight of 2 T. through 
10 ft., what part of its K.E. is lost ? iaoo ^ 

Ans. (a) foot-tons, (c) 20 foot-tons. 

3. A circular disk of mass wii is suspended by a horizontal axis passing 
through its center. A flexible thread is wound around its exterior and carries a 
mass m 2 attached to its free extremity. Show that the angular acceleration is 

— ~- — : , where r = radius of disk. What distance will m 2 fall from rest in t 

r(2m 2 + mi)' 

seconds ? An*, h = m "-^ 

m\ + 2 7i%2, 

4. A solid cylinder rolls down an inclined plane whose inclination is a. Show 
that the linear acceleration of the center is constant and equal to § <7sin a. 

5. Compare the time of descent of the rolling cylinder in problem 4 with that 
of a body which slides without friction. AnSt y'g . ^2. 

6. Two equal particles revolve in a horizontal plane around a vertical axis at 
distances a and b. At what distance from the axis must both particles be placed 
together in order that the K.E. may remain unchanged ? Ans. r 2 = h (a 2 + 6' 2 ). 

7. A solid fly wheel weighs Wlh. and makes 1ST revolutions per minute. Its 
radius is r ft. and that of the axle c in. If the frictional retarding force on the axle 
is F units per pound, find the number of revolutions before stopping. 

Ans. -^L. 
600 Fcg 

8. A sphere weighing 100 lb. rotates about a horizontal diameter, making 
80 revolutions per minute. Find the K.E. ^ 40 ^^ foot-pounds. 

g 

9. A solid sphere rolls down an inclined plane. Show that the acceleration 
of the center is constant and equal to f g sin a. 

10. Compare the times of descent of the cylinder (problem 4) and sphere 
(problem 9). Ans . Vl5:Vl4. 



DYNAMICS OF A RIGID BODY 



241 



11. A uniform rod of length 2 a turns about a screw as in the figure. How 
high will it rise if an angular velocity w about the axis I is imparted to it ? 

Ans. h = °^. 

12. Show that the acceleration of the center of a hoop rolling 
down a hill is \g sin a. 

13. Show that the acceleration of the center of a hollow 
sphere rolling down a hill is § g sin a. 

14. Compare the times of descent of a hollow and a solid 
sphere rolling down an inclined plane. Ans. 5 : V21. 

15. Two bicyclists, riding exactly similar machines, coast 
down a hill, starting with equal velocities at the top. Neglecting the forces of 
friction and the resistance of the air, show that the heavier rider will reach the 
bottom first. 




16. A straight piece of uniform wire is stood vertically on end and allowed to 
fall over. With what velocity does its extremity strike the ground ? 

Ans. VSlg. I = length of wire. 



17. A train of T tons descends an incline of s ft. in length having a total fall 
of h ft. What will be the velocity at the bottom, friction being p lb. per ton ? 

1 



Ans. v 2 = 2 gh 



1000 



gps. 



18. A uniform cylindrical rod 6 ft. long, radius 2 in., and density 5, is 
suspended so that it swings freely in a vertical plane about one end. It is dropped 
from a position making an angle of 30° with the vertical. Find the K.E. and an- 
gular velocity as it passes through the vertical position. 



Ans. K.E. 



5x 

2 



1 ] foot-pounds, 



1 -™ 



19. A homogeneous cylinder of mass 31 and radius a can turn around its axis, 
which is horizontal. A fine thread supporting a mass m is wound around it. Find 
the angular velocity of the cylinder when m has descended a distance h. 

Ans. * 2 = 4m -^ 

• a 2 (M + 2m) 

20. Show that a cylinder of altitude a and radius b rotating about its axis has 

enough energy to raise a weight equal to its own through a vertical distance — ^-. 

4g 

21. A sphere whose radius is a ft. rolls without sliding down an incline of 
30°. If the mass of the sphere is m, find its velocity after rolling a distance s. 
Find the time required to roll this distance. If the plane is 200 ft. long, what is the 
velocity of the center of the sphere at the bottom of the incline ? How far up an 
incline of 45° would it run ? A nQ „& — & — .*> 28 



Ans. 



p. 
= -gs 



t 2 = 



5<7 



t a = 1000y . 



229 ft. 

V2 



242 



THEORETICAL MECHANICS 




22. A bowling alley has a return trough 60 ft. long with a slope of 1 ft. in 15 ft. 
(a) Neglecting friction, discuss the motion of a ball 
whose radius is 4 in. and mass 8 lb. when allowed to roll 
from rest at the higher end. (6) Suppose the trough is 
triangular and the angle 90°. The ball then rolls on two 
points on its sides. Discuss the 
motion. 

Ans. («)/ = £,, ( & )/=^y 

23. Two masses mi and m 2 (wi>wi2) hang over a 
smooth pulley by means of a flexible, inelastic thread whose 
mass can be disregarded. Discuss the motion (a) leaving 

out the mass w 3 of the pulley, (&) taking 
account of the pulley's mass. 

24. A marble of radius a starting practically from rest at 
the upper end of the vertical diameter, ' rolls off a sphere of 
radius B. At what point will it cease to touch the sphere ? 

Ans. After the center of mass has descended a vertical 
7 
distance = — (B + a). 
17 

25. A uniform rod whose length is 2 b oscillates in a vertical plane about a 
horizontal axis distant a from its center of mass. Find the length of the equivalent 
simple pendulum. Find also the center of oscillation when the rod is suspended 
from one end. . 7 , & 2 ~ „ 4 , 





Ans. I 



3a 



OC 



26. A pendulum formed of a right circular cylinder of radius r and length h 
oscillates about a diameter of one of its bases as a fixed horizontal axis. Find the 
period. 



Ans. T=2tt 



V 



3r 2 + 4ft 2 

6^ 



27. Suppose the cylinder in problem 26 falls from the vertical position above 
the point of support. What is its angular velocity when it has turned through an 
angle of 6 = f v ? What should be its angular velocity at the lowest point in order 



that it may just rise to its original position ? . 2 «_ 18 hg 

W ~3r 2 + 4/i 2 



24 gh 



3 r 2 + 4 W- 



28. A circular lamina of radius r oscillates (a) about a tangent lying in its 
plane, (5) about a line through the circumference perpendicular to its plane. Find 
in each case the length of the equivalent simple pendulum. 

Ans. (a)l = lr; (6) l = %r. 

29. A cube whose edge is a swings as a pendulum about a horizontal edge. 
Find the length of the equivalent simple pendulum and the period. 

Ans. Z = |aV2. 



30. A circular arc oscillates about an axis through its middle point perpendic- 
ular to its plane. Show that the equivalent simple pendulum is independent of the 
length of the arc and equal to twice the radius. 



DYNAMICS OF A RIGID BODY 



243 



31. A given sphere has a radius of 6 in. If the axis of revolution is a hori- 
zontal tangent, find the moment of inertia and show that the length of the equiva- 
lent simple pendulum is 8.4 in. 

32. A right circular cone of height h and radius r oscillates about a horizontal 
axis perpendicular to its own axis at the vertex. Show that the length of the 

4 K 2 + f 2 



equivalent simple pendulum is 



5ft 



33. A square whose side is a oscillates about a horizontal axis perpendicular 

to its plane. How far from the center of the square is this axis when the period has 

a minimum value ? A a 

Ans. ■ — -• 

VB 

34. Find the axis about which an elliptic lamina must oscillate that the time of 
an oscillation may be a minimum. 

Ans. The axis must be parallel to the major axis and bisect the semiminor axis. 



35. A heavy disk weighing W lb. is set in motion by a 
weight P as indicated in the figure. The radius of the disk is a 
and that of the axle is b. The mass of the axle may be disregarded 
in comparison with that of the disk, (a) What is the angular 
velocity of the disk when P has descended h ft. ? What time is 
required ? 



36. A cord passes over a smooth peg as shown in the figure. 

To one end of the cord is attached a mass m which falls 

C~A op vertically, and the other end is fastened to the axle of 

a solid cylinder of mass M and radius B which rolls on 
a horizontal plane. Find the acceleration of m and the 




W////////////M 




frictional force. 



37. In problem 36 suppose M is a solid sphere. 



Ans. f=-™Z 



m+%M 



38. In problem 36 suppose 31 is a hollow sphere. 

39. A cord passes over a smooth peg as shown in the figure. To one end of 
the cord is attached a mass m which falls vertically and 
the other end is wrapped around a solid cylinder which 
rolls on a horizontal plane. Find the acceleration of m. V/Mffi/ 



Ans. f = 



mg 



m + lM 




40. In problem 39 suppose M is a hollow cylinder of negligible thickness. 




41. Suppose the cylinder of problem 36 rolls on a 

plane whose inclination is a. Find the acceleration of m 

and the frictional force. . (m — M sin a) a 

Ans. •* ^ . 

m + | M 

42. In problem 41 suppose M is a solid sphere. 



244 



THEORETICAL MECHANICS 



43. In problem 41 suppose M is a hollow sphere. 

44. Suppose the cylinder of problem 39 rolls on a plane 
whose inclination is a. Find the acceleration of to. 

Ans ( 2 m ~ M sin a ^ g 

45. In problem 44 suppose M is a hollow cylinder of 
negligible thickness. 

46. In problem 41 suppose to and M given. Determine the inclination a so 




that the acceleration of to is zero. 



Ans. sin a = 



M 

47. Two masses to and to' suspended from a wheel and axle do not balance. 
The radius of the wheel is «, and that of the axle is b. Show that the acceleration 

of to is ( wa ~ m b ) a 9 where i" is the moment of inertia of the machine about its 
ma 2 + m'b 2 + I 

axis. 

48. The handle of a wheel and axle is let go just as a bucket full of water 
weighing 60 lb. reaches the top of a well 18 ft. deep, and the bucket gets to the 
bottom again in 6 sec. If the axle is 6 in. in diameter, find the moment of inertia 
of the wheel and axle. Ans. 116.25. 

49. A prism whose cross section is a square, each side being a, and whose 
height is h, oscillates about one of its upper edges. Find the length of the equiva- 
lent simple pendulum. Ans. iyV 2 + h 2 . 

50. A uniform cylinder has coiled around its central section a light, perfectly 
flexible string. One end of the string is attached to a fixed point, and the cylinder 
is allowed to fall. Show that it will fall with acceleration f g. 

51. An elliptic lamina swings about a horizontal axis which passes through 
one focus, is perpendicular to the major axis, and lies in the plane of the ellipse. 
The other focus is the center of oscillation. Prove that the eccentricity is J. 

52. Two smooth planes are placed back to back, as 
shown in the figure. The body M 2 slides down the plane 
of inclination a 2 , and by means of a cord passing over a 
pulley of mass Mi draws the cylinder of mass M 3 up the 
plane of inclination «3. Supposing the cylinder rolls 
without slipping, determine the acceleration. , 2 g(M 2 sin a 2 




M 8 sin g s ) 



M 1 + 2M 2 + 3 Ms 



CHAPTER XIII 

EQUILIBRIUM OF COPLANAR FORCES 

113. Equilibrium of forces. If a system of forces acting upon 
a body produces no change of motion, the forces are said to be in 
equilibrium ; and if the body is initially at rest, it will remain at 
rest under the action of a system of forces in equilibrium. The 
part of mechanics which deals with systems of forces in equilib- 
rium is called statics. Assuming that the body is initially at rest, 
the problem of statics is the determination of the conditions upon 
the forces acting in order that the body shall remain at rest. 

114. Analytic conditions for equilibrium of coplanar forces. 

Consider a uniplanar motion with the following characteristics : 
(#) the center of gravity moves in a straight line with constant 
speed; (5) the angular velocity remains constant. Referring to 
the fundamental equations ((V), Chapter XII) 

M— = F , 

dP *' 



(1) 



M dP ~ ** 

T cPe_ T 



then by the hypothesis (a) J = g = 0, and by (5) g=0. 

Hence for the motion described 

(2) F x = 0,F y = 0,L=0, 

that is, the sum of the axial components of all forces acting is zero, 
and the resultant moment with respect to the origin of all forces act- 
ing also vanishes. 

Next assume a system of coplanar forces (that is, whose lines 
of action lie in a plane) such that equations (2) are satisfied. 

245 



246 THEORETICAL MECHANICS 

Suppose these forces act upon a rigid body whose previous motion 
(£ = 0) possessed the characteristics (a) and (6), assuming that the 
directing plane is parallel to the plane of the forces. Then, since 
by (1) we have to integrate 

d 2 x _ ^ d 2 y _ r, d 2 6 _ « 

it is clear that the center of gravity will continue to move uni- 
formly in a straight line, and further that the angular velocity 
will be unchanged. Hence the motion is entirely unchanged. It is 
also clear that such a system of forces will not disturb the body if 
it is initially jat rest. 

Hence the 

Theorem. A system of coplanar forces is in equilibrium if and 
only if (1) the sum of the X-eomponents of all the forces is zero, (2) 
the sum of the Y-components of all the forces is zero, (3) the sum of 
the moments ivith respect to the origin of all the forces is zero. 

If the forces be denoted by F v F v ••• F n , the angles which 
their lines of action make with the .X-axis by a v a 2 , ••• a n , and 
the lever arms with respect to the origin by d v d 2 , ••• d n , the con- 
ditions for equilibrium may be written 



(3) 



F x = F x cos a x -f F 2 cos a 2 -f- -•• F n cos a n — 0, 
F y — F x sin a x + F 2 sin a 2 + ••• F n sin a n = 0, 
I L = F x d, + F 2 d 2 + .- FJ n = 0. 



The first two equations are the conditions that the acceleration of 
the center of gravity shall be zero, and the third is the condition 
that the angular acceleration shall be zero. 

Since any point in the plane may be chosen for the origin and 
since any line through the origin may be taken for the X-axis, we 
see that if a system of forces is in equilibrium, (a) the sum of the 
components in any direction is zero, (5) the sum of the moments 
with respect to any point is zero. And conversely, if (a) and (6) 
are satisfied, the system is in equilibrium. 

From (3) we may deduce other forms of the conditions of 
equilibrium which are convenient in applications. 




EQUILIBRIUM OF COPLANAR FORCES 247 

I. A system of forces is in equilibrium if the sum of the compo- 
nents along any two intersecting {not coincident) straight lines is zero, 
and the sum of the moments with respect to one 
origin is zero. 

Proof The second part of I is the con- 
dition L = 0. To prove the first part, let the 
two lines be OX and OA, and denote the angle 
XOA by ft- The sum of the components 
along OX is 

(4) F 1 cos a 2 + F 2 cos « 2 + ••• F n cos a n = F x = 0. 
The sum of the components along OA is 

(5) F 1 cos (« : - ft) + _F 2 cos (« 2 — /3) + ••• -F„ cos (a n - #) = 0. 

Equation (5) may be written in the form 

F 1 cos aj_ cos ft + F x sin a x sin ft -\- F 2 cos a 2 cos ft -\- F 2 sin a 2 sin /3 
-f- • • • F n cos a n cos ft + i^ ?l sin «„ sin ft = 0. 

Since F x = 1 F { cos a„ F y =L F t sin «^, this equation becomes 

.F, cos /3 + ^ sin /3 = 0. 

Since by (4) i^ = and ft =£ by hypothesis, this equation gives 

^ = 0. 

Hence (4) and (5) are equivalent to the first two of equa- 
tions (3). Q.E.D. 

II. A system of forces is in equilibrium if the sum of the moments 
is zero for each of two origins, O and C, and the sum of the com- 
ponents is zero in any direction not perpendicu- 
lar to OC. 



(b,c) 



(6) 



Proof. Take the point O for origin of 
coordinates, and the X-axis parallel to the 
direction of resolution. 

]£ Let the point of application of F 1 be 

(x v y x ). Then (Art. 62), 

moment of F x with respect to = L 1 = 
x x F x sin a 1 — y 1 F 1 cos a v 



248 THEOEETICAL MECHANICS 

and similarly, 

(7) moment of F 1 with respect to C(b, c) == 

(x x — b)F 1 sin a x — (jy x — c)F x cos a x = 

L x — 5.F X sin a x + c^ cos a r 

Hence by summing up, if i = sum of moments with respect 
to 0, we shall have, using (3), 

(8) Sum of moments with respect to C= L — bF y + cF x . 

This vanishes by hypothesis. Also L—0, F x = by hypothesis. 
Hence F y =Q and (3) hold. q.e.d. 

III. A system of forces is in equilibrium if the sum of the 
Y moments is zero for each of three origins not 

on the same straight line. 

\fr cj Let the three centers of moments be (9(0, 0), 

A(a, 0), and (7(6, c). Then if the moments 
of the force F 1 are, respectively, L v L' v L'[, 



we shall have 



A X 

(a.o) . 

L x = x 1 k 1 sin a x — y l F 1 cos a v 

L[ = (x 1 — a^)F x sin a 1 — y l F 1 cos a x = L x — aF x sin a v 

U[ = (x x — b)F x sin a x — (^ — e)F 1 cos a x = L x — bF x sin a x + cF x cos a r 

Summing up for all the forces, and denoting these sums by Z, L\ 
U\ then, using (3), 

L' = L - aF y , L" = L- bF y + cF x . 

The hypothesis L = L' = L" = leads to the condition F x = 0, 

F y =Q. Q.E.D. 

In any problem of statics either I, II, or III may be used. 
The choice depends upon the convenience for the particular 
problem. 

A special condition which is important in the applications may 
be derived when the number of forces is three. Let the forces be 
denoted by F v F 2 , and F z . There are two cases to be considered. 

(1) Suppose the lines of action of F x and F 2 intersect in the 
point 0. Taking moments with respect to 0, we have 

moment of F x = moment of F 2 = 0, 
and, for equilibrium, 

moment of F 1 + moment of F 2 + moment of F z = 0. 



EQUILIBRIUM OF COPLANAR FORCES 249 

Hence moment of F z = 0, which means that the line of action of 
F 3 must pass through 0. The first two conditions of equilibrium 
(3) simply assert that the vector sum of the three forces is zero. 

(2) Suppose the lines of action of F 1 and F 2 are parallel. Let 
the axis OX be parallel to the lines of action of F x and F v Tak- 
ing components in the direction of the y-axis, we have 

and, for equilibrium, 

Hence F SlJ = 0, which means that F 3 is parallel to the .X-axis. 
Hence the 

Theorem. If three coplanar forces are in equilibrium, their 
lines of action are concurrent or parallel and their vector sum is zero. 

115. - General method of solving problems in equilibrium. The 
general problem of equilibrium of a system of forces is the follow- 
ing : given a body or system of bodies acted upon by a sj^stem of 
forces of which some are known, to determine the unknown forces 
so that the system is in equilibrium. For the solution we have 
the three conditions of equilibrium as expressed by I, II, or III 
and such geometric conditions as may be implied in the statement of 
the problem. Most problems of statics can be solved in different 
ways and the best method is to be found only by experience. 
The general method of procedure is indicated in the following 
steps : 

(1) Draw a figure showing the body acted upon and represent 
by vectors all the forces acting. 

(2) Enumerate all the forces acting, specifying the magnitude 
and direction of each so far as known. 

(3) Write the three conditions of equilibrium, using I, II, or 
III to make the equations as simple as possible. 

(4) If the equations of equilibrium are sufficient to determine 
the unknown quantities, solve them. 

(5) If not, write as many equations as possible from the 
geometric conditions. 

(6) If the problem is determinate, the number of static and 
geometric equations is sufficient to determine the unknown quanti- 
ties by algebraic solution. 



250 



THEOEETICAL MECHANICS 



A^ 







W / 




ILLUSTRATIVE EXAMPLES 

1. A heavy uniform rod, AB, is fastened at A with a smooth hinge and is sup- 
ported in a horizontal position by a string attached at P and making an angle a 
with the horizon. Determine the tension in the string and the magnitude and 
direction of the force exerted by the hinge. 

Solution. Following the steps indicated above we 
(1) draw the figure. 

(2) The forces acting on the rod are three : (i) the 
known weight W acting downwards at 0, the middle 
point of AB, (ii) the tension T of unknown magnitude, 
acting at B in a direction indicated by the angle a, (iii) the force P of the hinge at 
A, unknown in magnitude and direction. 

(3) Since the number of forces is three, we may conclude from the theorem of 
Art. 114 that DB and FA intersect in a point E vertically under 0. This deter- 
mines the direction of the force P, since the angle FAO is it — a. 

Resolving the forces in directions parallel and perpendicular to OP, we have 

Tcos a — Pcosa = 0, 
^ . Tsina + Psin a — W=0. 

The moment equation has been used in applying the theorem of Art. 114. 

W 

(4) The solution of equations (1) gives T= P = — cosec a. 




2. A uniform rod AB rests with the end A against the corner of a smooth * 
horizontal floor and a smooth vertical wall. At the end B two strings are attached 
of which one is fastened to a point C in the wall. The 
other passes over a smooth peg at D in the floor, making 
ABB a right angle, and supports a weight T. The weight 
of the rod is W pounds and the tension in PC is F pounds. 
Determine the weight T and the pressures at A. 

Solution. (1) In the figure the middle point of AB 
is the center of gravity of the rod, a is the inclination of 
AB to the horizon, and /3 is the inclination of BC to' the 
horizon. 

(2) The forces acting on the rod are five in number : (i) the weight W acting 
downwards at the middle point of AB, (ii) the known tension F in BC, (iii) the 
tension T in PP, unknown in magnitude, (iv) the pressure Pi of the vertical wall 
at A, unknown in magnitude, (v) the pressure P 2 of the floor at A, unknown in 
magnitude. 

(3) For the equations of equilibrium we shall use II. Resolving in a horizontal 
direction, 

(1) - F cos /3 + Pi + Tsin a = 0. 

Taking moments about A and denoting the length of the rod by I, 
- \ Wl cos a - Tl + Fl sin (« + P) = 0, 



* A surface is defined as smooth if it can exert pressure only in the direction of the 
normal to the surface. 



EQUILIBRIUM OF COPLANAR FORCES 251 

or, 

(2) -i Wcosa- P-f- Psin(« + /3) = 0. 

Taking moments about P, 

| Wl cos a — P 2 l cos a + P\l sin a = 0, 
or, 

(3) \ TFcos a — P 2 cos a + P x sin a = 0. 

(4) The three equations of equilibrium are sufficient to determine the three un- 
known quantities Pi, P 2 , and T. From (1), (2), and (3) we find 

T = Fs'm (a +/3) — $ TFcos a. 



P 2 = Pcos (a + |8) sina+ 1 JF(1 + sin 2 a). 

3. A light rigid rod rests partly within and partly without a hemispherical 
smooth bowl, which is fixed in space. A weight W is clamped on to the rod at a 
point C within the bowl. Determine the position of equilibrium and all forces act- 
ing on the rod. y % 

Solution. (1) The figure represents a section cut / ijVj ^ 

from the bowl by the vertical plane determined by the A -/- ib^fR 

center and the rod AF. The position of equilibrium \ J^^^vA JI 
is known if the inclination of AB to the horizontal is \j^£$L_ / 

known. ^ ^ s =^= sS ^^ 

(2) The forces acting on the rod are three in number : (i) the weight W acting 
downwards at C, (ii) the unknown pressure Pi of the surface of the bowl at A, 
acting in the direction of the normal, (iii) the unknown pressure P 2 of the edge of 
the bowl at P, acting in a direction perpendicular to AF. 

(3) For the equations of equilibrium we shall use I. Resolving in a horizontal 
direction, 

(1) - P 2 sin + Pi cos = 0. 
Resolving in a vertical direction, 

(2) P 2 cos + Pi sin - W = 0. 

Taking moments about A, denoting the known distance AC by I and the unknown 
distance AB by x, 

(3) - Wl cos d + P 2 x=0. 

(4) The unknown quantities in (1), (2), and (3) are Pi, P 2 , 0, 0, and x. Hence 
we require some geometric conditions. 

(5) Since the curve DAB is a semicircle, the normal at A passes through O 
and it follows that = 20. Also, since AB is a chord of the circle, 

AB = x = 2 r cos 0, 
where r denotes the radius. 

(6) Substituting the values of <p and x in (1), (2), and (3), respectively, we 
obtain 

(4) P 2 sin - Pi cos 2 = 0, 

(5) P 2 cos + Pi sin 2 0= IF, 

(6) 2 rPo - Wl = 0. 



252 THEORETICAL MECHANICS 

Solving (4), (5), and (6), we find 



8r 

Pi = JFtantf, P 2 = — . 
2r 

It may be remarked that the angle of equilibrium does not depend on the mag- 
nitude of the weight attached to the rod. 



PROBLEMS 

1. A rod AB is hinged at A and supported in a horizontal position by a 
string BC making an angle of 45° with the rod. A weight of 10 lb. is suspended 
from B and the weight of the rod may be neglected. Find the tension in the string 
and the force at the hinge. Ans. 10 V2 10 lb. 

2. A wheel capable of turning freely about a horizontal axis has a weight of 
2 lb. fixed to the end of a spoke which makes an angle of 60° with the horizon. 
What weight must be attached to the end of a horizontal spoke to prevent motion 
taking place ? 

3. Two weights, P and Q, rest on a smooth double inclined plane as shown in 
the figure, and are attached to the extremities of a string which passes over a smooth 
peg O at a point vertically over the intersection of the planes, the peg and the 
weights being in a vertical plane. Find the position of equilibrium. 

A Ans. The position of equilibrium is given by the equations 

p sin g __ q sinff 
cos 6 cos 0' 

cos a cos g _ I 
sin 6 sin h 1 
where I is the length of the string and h= CO. 

4. A bar of mass 15 lb. , whose center of gravity is at its middle point, rests 
with its ends upon two smooth planes inclined to the horizon at angles of 36° and 
45° respectively. Determine the inclination of the bar to the horizon when in 
equilibrium, and the pressures exerted upon it by the supporting planes. 

Ans. 10° 39', 8.93 lb., 10.74 lb. 

5. A uniform rod 15 in. long and weighing 12 lb. has a weight of 10 lb. 
suspended from one end. At what point must the rod be supported that it may just 
balance ? Ans. 4^ in. from the weight. 

6. Prove that three forces acting at the middle points of the sides of a triangle 
perpendicularly inwards, and proportional to the lengths of the sides, are in 
equilibrium. 

7. Extend the theorem of problem 6 to a plane polygon of any number of 
sides. 

8. ABCD is a plane quadrilateral, P and Q are the middle points of the oppo- 
site sides AB and CD, and O is the middle point of PQ. Prove that the four forces 
represented by OA, OB, OC, OD, respectively, are in equilibrium. 




EQUILIBRIUM OF COPLANAR FORCES 253 

9. A uniform beam of weight W and length 3 ft. rests in equilibrium with its 
upper end A against a smooth vertical wall, while its lower»end B is supported by a 
string, 5 ft. long, whose other end is attached to a point C in the wall. Find AC 

and the tension in the string. , 4 Wq 

Ans. AC=— ft., T= ~ W. 
V3 8 

10. ABCD is a plane quadrilateral. Forces act along the sides AB, BC, CD, 
DA, measured by a, |3, 7, 5 times those sides respectively. Show that if these forces 
are in equilibrium, then ay = /35 . 

11. A bar AB, whose center of gravity is at its middle point and whose mass 
is 12 lb., is supported in a horizontal position by strings attached to the ends, and 
sustains loads of 16 lb. and 20 lb. at A and B respectively.* If the string at A is in- 
clined 45° to the horizon, what is the inclination of the string B ? Find the tensions 
in the strings. Ans . 490 47/^ 31>12 lb., 34.05 lb. 

12. Three smooth pegs A, B, C stuck in a wall are the vertices of an equilateral 
triangle, A being the highest and the side BC horizontal. A light string passes once 
around the pegs and its ends are fastened to a weight W which hangs in equilibrium 
below BC. Find the pressure on each peg. 

13. A weightless string is suspended from two fixed points and at given points 
on the string equal weights are attached. Prove that the tangents of the inclina- 
tions to the horizon of different portions of the string form an arithmetic progression. 

14. A uniform yardstick weighing 10 oz. is supported in a horizontal position 
by the thumb at one end and the forefinger at a point 3 in. from the end. What is 
the pressure on the thumb and on the finger ? ^ nSi 59 oz# 60 oz. 

15. A beam AB weighing \ T. per running foot and 18 ft. long is loaded with 
4 T. at A and 5 T. at B. It is supported at points 4 ft. from A and 6 ft. from B. 
Find the supporting forces P and Q. AnSt p = 55 x , Q = 12^ T. 

16. A uniform rod of weight 50 lb. and length 18 ft. is carried on the shoulders 
of two men who walk at distances of 2 and 3 ft., respectively, from the two ends. A 
weight of 50 lb. is suspended from the middle point of the rod. Find the total 
weight carried by each man. 

17. A uniform plank 20 ft. long, weighing 42 lb., is placed over a rail, and two 
boys weighing 75 and 99 lb., respectively, stand each at a distance of 1 ft. from each 
end. Find the position of the plank for equilibrium. 

Ans. 1 ft. from the middle point. 

18. Two equal weights, P, Q, are connected by a string which passes over two 

smooth pegs A, B, situated in a horizontal line, and supports a weight W which 

hangs from a smooth ring through which the string passes. Find the position of 

equilibrium. „ , , . W AT > 

Ans. The depth of the ring below the line ^4P» is — . — * <&■&. 

F ° 2\/4P^-JF2 

19. A light rod rests wholly inside a smooth hemispherical bowl whose radius 
is r, and a weight W is clamped on to the rod at a point whose distances from the 
ends are a and b. Show that, if 6 be the inclination of the rod to the horizon in the 
position of equilibrium, then 

sinfl= a ~ h . 



2vV 2 - ab 



254 THEOEETICAL MECHANICS 

20. Two weights, P and Q, rest on the concave side of a parabola whose axis is 
horizontal, as shown in trhe figure, and are connected by a light string, of length Z, 
which passes over a smooth peg at the focus F. Find the posi- 
tion of equilibrium. , 

Ans. If is the angle which FP makes with the axis, and 
^ "?fl ' 4 m is the latus rectum of the parabola, then 




PVl-2m 

COt a — 



2 Vm(P 2 +# 2 ) 

21. In problem 20 show that the depths of the weights below the axis are pro- 
portional to their masses.* 

22. A particle is placed on the convex side of a smooth ellipse, and is acted 
upon by two forces F and F', towards the foci, and a force F n , towards the center. 
Find the position of equilibrium. 

Ans. r = — , where r is the distance of the particle from the center of the 

Vl - rfi 

ellipse, b is the semiminor axis, and n = 

116. Friction. A smooth surface is defined as one which can 
exert upon a body in contact with it only a pressure in a direc- 
tion normal to the surface. Such surfaces do not exist in nature. 
Suppose a heavy box is at rest upon a horizontal table. If the 
table were smooth, the box could be moved by any horizontal 
pull, the acceleration being, by Newton's Second Law of Motion, 
directly proportional to the force and inversely proportional to 
the mass of the box. Experiment shows, however, that this is 
not true. If the horizontal pull is slight, no motion ensues, and 
consequently the forces acting on the box are in equilibrium. 
The forces acting are three in number : (i) the weight acting 
vertically downwards, (ii) the horizontal pull H, and (iii) the 
pressure of the table. From the principles of equilibrium it 
follows that the pressure of the table must be made up of two 
components, of which one, numerically equal to the weight, acts 
vertically upwards, while the other is numerically equal to H but 
opposite in direction. A rough surface can exert upon bodies in 
contact with it a pressure made up of (1) a component normal 
to the surface called the normal pressure, and (2) a component 
tangent to the surface called the friction. All physical surfaces 
are more or less rough. 

Our knowledge of frictional forces is obtained by experiment 
and is expressed in the following 



EQUILIBRIUM OF COPLANAR FORCES 



255 



Laws of Friction. 1. If the body is in equilibrium, the fric- 
tion is equal and opposite to the tangential component of the applied 
forces. In the preceding example there is no friction if there is 
no horizontal pull. 

2. No more than a certain amount of friction can be called into 
play. The value of the friction when sliding is just about to take 
place is called the limiting friction. 

3. The magnitude of the limiting friction bears a constant ratio 
to the normal pressure. This constant ratio, /jl, is called the 
coefficient of friction, and its value depends upon the nature of 
the surfaces in contact. 

4. The coefficient of friction is independent of the area of contact 
of the two bodies if the touching surfaces are uniform in character. 

The angle of friction, \, is defined by (Art. 67) 



tan \ = fji. 



The coefficient of friction for various substances has been 
determined by experiment and some of the results 'are given in 
the following table of values for /jl 

Wood on wood, dry . 
Wood on wood, soaped 



Metals on oak, dry 

Metals on oak, wet 

Metals on oak, soaped 

Leather on oak, wet or dry 

Metals on metals, dry 

Metals on metals, wet 

Smooth surfaces, occasionally lubricated 

Smooth surfaces, thoroughly lubricated 



0.25 to 0.5 

0.2 

0.5 to 0.6 

0.24 to 0.26 

0.2 

0.27 to 0.35 

0.15 to 0.2 

0.3 

0.07 to 0.08 

0.03 to 0.036 



The values of //, given above are the coefficients of static fric- 
tion as defined in 3. The coefficient of dynamic friction (see 
Art. 67) is slightly less in numerical value than the coefficient of 
static friction. 

Illustrative Example. A uniform rod rests with one end on a rough hori- 
zontal floor and the other against a rough vertical wall. Supposing the coefficient 
of friction to be the same at both ends, determine the least inclination it can make 
with the horizon. 



256 THEORETICAL MECHANICS 

Solution. (1) In the figure, AB is the rod which is just about to slide down. 

(2) The forces acting on the rod are five in number : (i) the known weight W 

acting vertically downwards at C, (ii) the normal pressure B of the floor, unknown 

in magnitude, acting vertically upwards at A, (iii) the friction at A, of magnitude 

fiB and acting in the direction AD, (iv) the normal pressure 

B' of the wall at B, unknown in magnitude, (v) the friction 

l*B' at B acting vertically upwards. 

(3) Eor the conditions of equilibrium we shall use I, 
Art. 114. Resolving in a horizontal direction, 

(1) B'=fxB. 
Resolving in a vertical direction, 

(2) B + fxB' = W. 

Taking moments about A, and denoting the length of the rod by I, 

i Wl cos — fiB'l cos -B'l sin = 0, 
or, 

(3) i Wcos0 = B'(/x cos 6 + sine). 

(4) From equations (1), (2), and (3) we eliminate B and B', and solve for 0. 

The result is i _ „2 

tan0 = ± — ^. 




PROBLEMS 

1. A body of mass W pounds is at rest upon a plane making an angle with 
the horizon. A cord attached to this body runs parallel to the plane, passes over a 
smooth pulley, and sustains a weight of P pounds. Determine the magnitude and 
direction of the friction, the normal pressure, and the total pressure exerted by the 
plane upon the body. 

2. In problem 1 let W= 50 lb., P= 40 lb., = 32°, and suppose the body is 
just about to slide up the plane. Determine the coefficient of friction. 

Ans. m = 0.318. 

3. The roughness of a plane of inclination a is such that a body of mass W 
can rest on it. Find the least force required to draw the body up the plane. 

Ans. W sin 2 a, inclined at an angle a to the plane. 

4. A uniform beam rests with one end on a rough horizontal plane and the 

other against a rough vertical wall, and, when inclined to the horizon at an angle 

of 30°, is on the point of slipping down. Supposing that the surfaces are equally 

rough, determine the coefficient of friction. . J_. 

° ' Ans. ix = ~^ 

V3 

5. A body of 30 lb. mass, resting on a plane inclined 45° to the horizon, is pulled 
horizontally by a force P. If the coefficient of friction is 0.2, between what limits 
may the value of P vary and still permit the body to remain at rest ? 

Ans. 20 and 45 lb. 

6. On a rough plane of inclination the greatest value of the force acting along 
the plane and producing equilibrium is double the least. What is the coefficient of 
friction? Ans , « = Uan0. 



EQUILIBRIUM OF COPLANAR FORCES 257 

7. If the angle of friction is 30°, what is the least force which will sustain a 
weight of 100 lb. on a plane whose inclination is 60° ? Ans. 50 lb. 

8. A ladder inclined at an angle of 60° to the horizon rests with one end on a 

rough pavement and the other against a smooth vertical wall. The ladder begins 

to slide down when a weight is put at its middle point. Show that the coefficient of 

* . . . V3 

friction is 

6 

9. A uniform ladder weighing 100 lb. and 50 ft. long rests against a rough 
vertical wall and a rough horizontal plane, making an angle of 45° with each. If 
the coefficient of friction at each end is f , how far up the ladder can a man weigh- 
ing 200 lb. ascend before the ladder begins to slip ? Ans. 47 ft. 

10. A heavy body is placed on a rough plane whose inclination to the horizon 
is arc sin |, and is connected by a string passing over a smooth pulley with a body 
of equal weight which hangs freely. Supposing that motion is on the point of ensu- 
ing up the plane, find the inclination of the string to the plane, the coefficient of 
friction being 0.5. Ans. 6 — 2 arc tan \. 

11. Two weights rest on a rough inclined plane and are connected by a string 
which passes over a smooth peg in the plane. If the angle of inclination a is greater 
than the angle of friction e, show that the least ratio of the less to the greater is 
sin (a — e)/sin (a + e). 

12. Two equal weights are attached to a string laid over the top of two inclined 
planes, having the same altitude, and placed back to back, the angles of inclination 
of the planes being 30° and 60° respectively. Show that the weights will be on the 
point of moving if the coefficient of friction between each plane and weight be 

1 

2+ V3 

13. A body is supported on a rough inclined plane by a force acting along it. 
If the least magnitude of the force, when the plane is inclined at an angle a to the 
horizon, be equal to the greatest magnitude when the plane is inclined at an angle 
/3, show that the angle of friction is \{a — /3). 

14. A cubical block rests on a rough plank with its edges parallel to the edges 
of the plank. If, as the plank is gradually raised, the block turns over on it before 
slipping, what is the least value of the coefficient of friction ? 

15. It is observed that a body whose weight is known to be W can be just sus- 
tained on a rough inclined plane by a horizontal force P, and that it can also be 
just sustained on the same plane by a force Q up the plane. Express the angle of 
friction in terms of these known forces. , x _ arc CQg PW 



QVP 2 + W 2 

16. It is observed that a force Qi acting up a rough inclined plane will just 
sustain on it a body of weight W, and that a force Q 2 acting up the plane will just 
drag the same body up. Find the angle of friction. 

->. sin # 2 ~ 9l 



Ans. X 



2VW 2 - Q1Q2 



258 THEORETICAL MECHANICS 

17. A heavy uniform rod rests with its extremities on the interior of a rough 
vertical circle. Find the limiting position of equilibrium. 

Ans. If 2 a is the angle subtended at the center by the rod, and \ the angle of 

friction, the limiting inclination of the rod to the horizon is given by the equation 

n sin 2 X 

tan 6 = 

cos 2 X + cos 2 a 

18. An insect tries to crawl up the inside of a hemispherical bowl of radius a. 
How high can it get if the coefficient of friction between its feet and the bowl is % ? 

19. Two equal rings of weight W are movable along a curtain pole, the coeffi- 
cient of friction being \x. The rings are connected by a loose string of length Z, which 
supports by means of a smooth ring a weight W\. How far apart must the rings be 
so that they will not come together ? 

117. Equilibrium of flexible cords. It is assumed that the 
cords discussed in tins article are inextensible and perfectly 
flexible. The cross section is supposed to be small so that we 
may consider the curve formed by the cord. For a perfectly 
flexible cord in equilibrium it is evident that the resultant force 
at any point must act in the direction of the tangent to the curve 
formed by the cord. We wish to investigate the form of the 
curve assumed by a cord which is fastened at both ends and which 
sustains a weight distributed according to a given law. Since 
the cord is in equilibrium it is evident that, if any segment be 
replaced by a rigid wire of the same shape and bearing the same 
load, the system would still be in equilibrium. In order to deter- 
mine the form of the curve we may 
consider any segment and treat it as a 
rigid body. 

Let the plane of the cord be the XY- 
plane with the !F-axis directed vertically 
upwards, and let co be a function (of the 
coordinates or length of arc) represent- 
ing the distribution of weight along the 
cord. Consider any segment P^Pt 
This segment is in equilibrium under the action of three forces : 
(i) the tension T v directed along the tangent to the curve at P x ; 
(ii) the tension T 2 , directed along the tangent at P 2 ; (iii) the 
weight W acting vertically downwards at (7, the center of gravity 
of the load of the segment. The weight W along the segment 
P X P 2 is the difference of values of the function co at P 2 and P v 
that is W = co — go*. 





Y 










V 


jrr 2 






cJ 








3^7\ 






^ 


/\2 




^ 





/ 


X 



EQUILIBRIUM OF COPLANAR FORCES 259 

Let fa and fa denote the inclinations of the tangents to the curve 
at P x and P 2 respectively. Resolving in a horizontal direction, 

(1) T x cos fa = T 2 cos fa. 
Resolving in a vertical direction, 

(2) T 2 sin fa = T x sin fa + W. 

Since P x and P 2 are any points on the curve, equation (1) 
shows that the horizontal component of the tension is the same at 
every point of the curve, and this is evidently equal to the tension 
at the lowest point. Denoting the constant horizontal component 
of the tension by H, we have, from (1), 

rp _ H m _ H 



COS (ftj COS fa 

With these values of T x and T 2 we may write equation (2) in 
the form 

(3) tan^-tan^ 1 =J= 6, 2^i. 

The function to is supposed to be known for every point of the 
curve. If H and the slope <f> 1 at some one point J > 1 has been 
determined, equation (3) may be used to determine the slope fa 
at any second point P 2 . 

In order to determine the shape of the curve we must find the 
differential equation which characterizes it. Let s denote the 
length of arc measured from some fixed point on the curve, s 1 and 
s 2 being the distances to P 1 and P 2 respectively. Dividing both 
members of equation (3) by s 2 — s v we have 



oo 



tan fa — tan fa _ 1 a> 2 — co^ 



h ~ h H h 



Now let P 2 approach P 1 along the curve. Then s 2 — s 1 approaches 
zero as a limit, the first member of equation (4) approaches 

^ , ^ , and the second member approaches — — - . Hence the 
as H ds 

differential equation of the curve is 

^r^ d (tan fa) _ 1 dco 

^ } ds ~ H ds' 

When co is given, the ordinary equation of the curve is found 
by integrating equation (5) and determining the constants by 



260 



THEORETICAL MECHANICS 



means of the initial conditions. In the following articles we con- 
sider the two cases which are most important in applications. 

118. The common catenary. The curve assumed by a heavy 
cord or by a cord carrying a weight distributed uniformly along 
the cord, is called a catenary. If w denotes the weight supported 
by unit length of the cord, then 

co = ws, 

and the differential equation of the curve [(5), Art. 117] becomes 

d (tan <fr) _ w_ 
ds " K 



(1) 



Since tan <j> is the slope of the curve we have (Calculus, p. 86) 



tan </> = 



dy 
dx 



Now, 



d_fdy s 
d fdy\ __ d fdy\ dx _ dx \dx, 
ds \dx) dx \dxj ds /~ /dy\ 2 



v 



1 + 



(Calculus, p. 142) 



dx) 



Writing -^ = p, and — = -, equation (1) becomes 

(XX XI c 



(2) 
Integrating, 
(S) 



dp _ dx 

c 



Vi +p 2 

log q> + vr+7 2 )=-+* r 



To determine the constant of integration e v we select the axes 
so that the Z-axis passes through (7, the lowest point of the cate- 
nary. The distance of the origin below O will 
be determined later. Since the tangent to the 

curve at C is horizontal, -&. = p = 0, when 

dx 

x = 0. Hence c x = 0, and equation (3) may 

x be written in the form 




Solving for p, 



P 



p-\- Vl -fp 2 = e 
i(e 



dy 
dx 



X 



EQUILIBRIUM OF COPLANAR FORCES 261 

Integrating, 

We now determine the distance 00 so that the constant of 
integration c 2 is zero. If, when x — 0, y — 00 — c, then c 2 = 0. 
Hence the 

Theobem. The equation of the catenary is 

(4) y = l(e f +e~°), 

where c, the intercept on the Y-axis, is the ratio of the horizontal 
tension to the weight per unit leyigth. 

119. Load distributed uniformly along the horizontal. This is 
the case of the cables supporting a suspension bridge if the weight 
of the cables is neglected in comparison with that of the bridge. 
When a cord supports a load distributed uniformly along the 
horizontal, the weight supported by any segment is proportional 
to the length of the projection of the segment on a horizontal 
line. If w' denotes the weight per horizontal unit, then 

a) = w'x, 

and the differential equation of the curve [(5), Art. 117] becomes 

.-j n d (tan <£) _ d fdy\ dx _ w' dx 

ds dx\dx) ds Hds 

Setting — = c ! , equation (1) takes the form 

(2) <^ = c'. 

K } dx* 

In order to determine the constants of integration, we choose 
the origin at the lowest point of the curve. The initial condi- 
tions are y = 0, -^ = 0, when x = 0. Integrat- 
ed 

ing (2) and imposing these conditions, we find 
for the equation of the curve 

(3) y = e ^x\ 

This equation represents a parabola with its axis vertical and 
latus rectum = \c'. Hence the 




262 THEOEETICAL MECHANICS 

Theorem. The curve assumed by a cord carrying a weight dis- 
tributed uniformly among the horizontal is a parabola with its axis 
vertical and latus rectum equal to \c\ where c' is the ratio of the weight 
per unit horizontal distance to the horizontal tension. 

PROBLEMS 

1. If s denotes the arc of the catenary measured from the lowest point, and <f> 
is the inclination of the tangent to the horizon, prove the following relations: 

(a) s = c tan 0, 
(6) y = c sec 0, 
(c) y 2 -s 2 =c' 2 . 

2. A cord hanging in the form of a catenary [(4), Art. 118] sustains a load of 
50 lb. per foot, and the tension at the lowest point is 1000 lb. The points of sus- 
pension are in the same horizontal line 100 ft. apart. Find -(a) the coordinates of 
the points of suspension, (5) the length of the cord, (c) the direction of the cord 
at the points of suspension. AnSt ^ ( ± 5 q 5 122.6), (5) 241.8, = 80° 37'. 

3. A uniform measuring chain of length I is tightly stretched over a fiver, the 
middle point just touching the surface of the water, while each of. the extremities 
has an elevation k above the surface. Show that the difference between the length 

Q i.2 

of the measuring chain and the breadth of the river is nearly — -. 

4. A chain 110 ft. long is suspended from two points in the same horizontal 
plane, 108 ft. apart. Show that the tension at the lowest point is nearly 1.477 
times the weight of the chain. 

5. A heavy chain hangs over two smooth fixed pegs. The two ends of the 
chain are free and the central portion hangs in a catenary [(4), Art. 118]. Show 
that the free ends are on the X-axis. 

6. A heavy uniform chain is suspended from two fixed points A and B in the 

same horizontal line, and the tangent at A makes an angle of 45° with the horizon. 

Prove that the depth of the lowest point of the chain below AB is to the length of 

V2 — 1 
the chain as . 

2 

7. If a and /3 are the angles which a uniform heavy string of length I makes 
with the vertical at the points of support show that the height of one point above 

the ° ther 1S Z COS !(<*+£) 

cos \ (a - /3) ' 

120. Stability. Suppose a heavy bead is constrained to slide 
on a wire in the form of a vertical circle. If the bead is at rest 
at the highest point A or at the lowest point B of the circle, the 
forces acting upon it are in equilibrium. If the bead is given a 
small displacement from the highest point A, the forces are no 
longer in equilibrium and the bead will move away from A. The 



EQUILIBRIUM OF COPLANAR FORCES 263 

position A is said to be a position of unstable equilibrium. If the 
bead is given a small displacement from the lowest point B, the 
forces are no longer in equilibrium, but the bead will return to its 
original position, and, if the wire is smooth, will perform small 
oscillations about B. The position B is said to be a position of 
stable equilibrium. 

Suppose the bead is constrained to slide on a horizontal wire. 
At any point on the wire the forces acting on the bead' are in 
equilibrium. If the bead is given a small displacement from a 
position P, it will remain in the new position. The position P is 
said to be a position of neutral equilibrium. 

To derive the analytic conditions for stability we make use 
of the potential function (Chapter X). Suppose a particle is con- 
strained to move without friction along a given path of any shape 
in a plane conservative field of force. Assume that the poten- 
tial U is a known function of the coordinates x and y. In addi- 
tion to the force of the field the particle is acted upon by a force 
of constraint (the pressure of the path) which, at any point of 
the path, is equal in magnitude but opposite in direction to the 
normal component of the force of the field. The resultant force 
acting on the particle is therefore the tangential component of the 
force of the field, and this is (Art. 94) 

F t = — = -(f x — +F^ 

ds \ x ds y ds. 

The necessary and sufficient condition that any position A (x v y-^) 
on the path shall be a position of equilibrium, is that 

But this is the condition (Calculus, p. 118) that the function XT 
shall be a maximum or minimum.* Hence the 

Theorem. For a position of equilibrium of a particle in a con- 
servative field of force, the potential energy is either a maximum or a 
minimum. 

* It may happen that the graph of the function U(s) has a point of inflection for the 
value s = s 1} corresponding to the point A (xi, y{) of the path of the particle. For ex- 
ample, suppose a heavy bead slides on a smooth wire in a vertical plane, and that the 
point A is a point of inflection where the tangent is horizontal. The point A is then a 
position of equilibrium for the bead, but the potential function U is neither a maximum 
nor a minimum. This special case is excluded in the statement of the theorems which 
follow. 



264 THEORETICAL MECHANICS 

The point A is a position of stable equilibrium if, when given 
a small displacement from A in either direction, the particle tends 
to return to A. From Art. 94 the force at any point in a con- 
servative field is directed towards the region of lower potential. 
Hence, if the particle returns to A, we may conclude that the 
value of the potential at A is smaller than at neighboring points 
of the path. In other words, at a position of stable equilibrium 
the potential function is a minimum. Similarly, at a position of 
unstable equilibrium the potential function is a maximum. 

Theorem. For a position of stable (unstable) equilibrium of a 
particle in a conservative field of force, the potential energy is a 
minimum (maximum ) . 

Since maximum and minimum values of a continuous function 
of one variable occur alternately, we have the 

Theorem. Along any given path in a conservative field of 
force, positions of stable and unstable equilibrium occur alternately. 

In solving problems to find the positions of equilibrium we 
may either (1) express the potential in terms of s, the length of 

arc along the curve, and use the condition — — = 0; or (2) we 

as 

may find the components of force and use the condition 

F x — -f- F v -^ — where y is expressed in terms of x by the equa- 
ls ds 

tion of the given curve; or (3) we may choose the direction along 
the curve so that s is an increasing function of x (or y), and 
examine the conditions under which 27", as a function of x (or y), 
is a maximum or minimum. 

ILLUSTRATIVE EXAMPLES 

1. A bead of mass m is constrained to move on a smooth curve y =f(x) in a 
field of force of which the potential function is U = — \ mu> 2 x 2 + mgy. Find the 
positions of equilibrium.* 

* This problem in the plane is equivalent to the following : 

A heavy bead slides on a wire in the form of the curve y = f (x), the F-axis being 
directed vertically upwards. The plane of the wire rotates about the F-axis with con- 
stant angular velocity o>. Determine the position of equilibrium of the bead. 

Suppose the bead is in a position of equilibrium A(x, y). It then revolves around 
the F-axis in a circle of radius x with angular velocity w. The bead exerts a horizontal 

pressure on the wire equal to ^- = mw 2 x (Art. 54). For motion of the bead along the 
wire the resultant of the forces acting is equivalent to that of the field specified above. 



EQUILIBRIUM OF COPLANAK FOKCES 265 

Solution. The components of the force acting at any point of the field are 
given by (Art. 91) f ^ n 

F x = — ^-^ = mw 2 x, 

BU 
F y =-<L" = -mg. 

For a position of equilibrium we have 

w dx „ dy _ n 
ds ds 



dx 



dy 



where ™ is found from the equation of the curve. 
dx 

Substituting the values from (1), the positions of equilibrium are found by 

solving the equations 

J* -'2- 

(2) \ 

2. Suppose the curve of example 1 is the straight line y = xtan a. Find the 
position of equilibrium and determine whether it is stable or unstable. 
Solution. Substituting in equations (2) we have 

ta 2 x = g tan a, 
y = x tan a. 
This set of equations has one solution, namely, 

x = ■&- tan «, y = — tan 2 a. 

W 2 CO 1 

To determine whether the equilibrium is stable or unstable, we express the 
potential in terms of the length of arc measured from the origin. Since the curve 
is a straight line, we have 

x = s cos a, y = s sin a. 
Hence 

U= — \ mofis 2 cos 2 a + mgs sin a. 
Differentiating, 

~y~ — — m(xfis cos 2 a + mg sin a, 
ds 

C ^E = - wa> 2 cos 2 cc. 
ds 2 

Since the second derivative of Uis negative, the function U is (Calculus, p. 124) 
a maximum, and the position is one of unstable equilibrium. 

PROBLEMS 

1. Suppose the curve of illustrative example 1 is the parabola x 2 = 2py. Show 
that there is no position of neutral equilibrium unless g =pu> 2 . If this condition is 
satisfied, then every point on the curve is a position of equilibrium. 

2. If the curve is the circle x 2 -t- y 2 = a 2 , find the position of equilibrium. 

Ans. y =— 9-. 



266 THEOKETICAL MECHANICS 

3. Suppose the curve is the cubical parabola 3 y = x z . Find the position of 

2 6 

equilibrium and prove that it is stable. Ans. x = — , y — — . 

g %3 

4. Suppose the curve is the semicubical parabola 9y 2 = 4x 3 . Find the position 
of equilibrium and prove that it is unstable. Ans. x = $-, y 



9 2 2 f/ 3 



5. A heavy bead slides on a smooth wire of any shape in a vertical plane. 
Discuss the positions of equilibrium. 

6. A unit particle is constrained to move along the curve ny = x n , in the field 
of force of which the potential function is U = — w 2 x 2 + 2gy. Show that the position 
of equilibrium is unstable, neutral, or stable according as n is less than, equal to, or 
greater than 2. 

7. A unit particle is constrained to move along the circle x 2 + y 2 = a 2 , in the 
field of force of which the potential function is U = Ax 2 + By 2 . Discuss the posi- 
tions of equilibrium. 

x 2 v 2 

8. A unit particle is constrained to move along the ellipse (- ^- = 1, in the 

field of force of which the potential function is U = x 2 + y 2 . Discuss the positions 
of equilibrium. 

9. In the preceding problem suppose the potential of the field is U = . 

Vx 2 + y 2 



CHAPTER XIV 

COLLECTION OF FORMULAS 

For the convenience of the student we give the following list 
of elementary formulas from Algebra, Geometry, Trigonometry, 
Analytic Geometry, and Calculus. 

FORMULAS FROM ALGEBRA 

1. Binomial Theorem (n being a positive integer): 

{a + by = a n + na^b + ^"^ a n ~ 2 b 2 + n(n- 1 ')(n-2) an _ %5 + _ 

[2 |_3 

+ nQ-l)Q-2)...(M-r + 2) ^_ r+1 ^._ 1 + . . ^ 

|r— 1 
Also written: 

(a + by = a n + {f\a n ~ l b + ("V - ^ + ("V" 8 ^ + - 

2. w! = [w = 1.2.3-4...(w-l>. 

3. In the quadratic equation ax 2 + bx+ c = 0, 

when J 2 — 4 ac > 0, the roots are real and unequal ; 
when b 2 — 4 ac = 0, the roots are real and equal; 
when b 2 — 4 ac < 0, the roots are imaginary. 

4. When a quadratic equation is reduced to the form 
x +px= q, p _ gum Q £ roo ^ s w ith sign changed, 

and q = product of roots with sign changed. 

5. In an arithmetic series, 

I = a + (n _ 1) d ; « = | (a + = | [ 2 * + O ~ 1) <*] • 

6. In a geometric series, 

7 „_i rl — a a(r n — l) 

r — 1 r — 1 

267 



268 THEORETICAL MECHANICS 

7. log ah = log a + log b, 9. log a n = nloga. 

8. log^ = loga-logb. io. logVa = -loga.' 

11. log 1 = 0. 12. log a a=l. 13. logi = -loga. 

(X 

FORMULAS FROM GEOMETRY 

14. Circumference of circle = 2 irr* 

15. Area of circle = irr 2 . 16. Volume of prism = Ba. 

17. Volume of pyramid = J Ba. 

18. Volume of right circular cylinder = irr 2 a. 

19. Lateral surface of right circular cylinder = 2 irra. 

20. Total surface of right circular cylinder = 2rrr(r + #)• 

21. Volume of right circular cone = ^7rr 2 a. 

22. Lateral surface of right circular cone = irrs. 

23. Total surface of right circular cone = irr(r -f- s). 

24. Volume of sphere = ^irr z . 25. Surface of sphere = 4 irr 2 . 

FORMULAS FROM TRIGONOMETRY 

26. sin x — ; cos x = ; tan x = 



csc# sec x cot# 

ow , sin x , cos x 

27. tan x = ; cot x — — 

cos x sm x 

28. sin 2 2; + cos 2 2; = 1 ; 1 + tan 2 x = sec 2 x ; 1 + cot 2 x — esc 2 x. 

29. sin x — cos ( - — x ) ; 30. sin (7r — #) = sin x ; 



,2 

cos x = sm — 



— — x); cos (7r — x) — — cos x ; 

tan 2; = cot f — — # J. tan (ir — x) = — tan 2;. 

31. sin (x + ^/) = sin x cos 2/ + cos x sin ?/. 

32. sin (x — y) — sin # cos ?/ — cos a; sin y. 

33. cos (:r + ?/) = cos 2; cos y — sin # sin y. 

* In formulas 14-25 r denotes radius, a altitude, B area of base, and s slant height. 



COLLECTION OF FORMULAS 269 

34. cos (x — y) = cos x cos y + sin x sin y. 

35. tan(> + y)= tana! + tan y - 

1 — tan a; tan 3/ 

36. tan( g -y) = , tana; - tal ^ 

1 4- tan a; tan y 

37. sin 2 a? = 2 sin a? eos #; , 9 „ 2 tan a? 



tan z# 



cos 2 a; = cos 2 a: — sin 2 a?; 1 — tan 2 x 

38. sina;= 2 sin-cos-; 2tan- 

2 2, 2 

tan a; = . 



cos x = cos 2 - — sin 2 — ; 1 — tan 2 - 

A A A 

39. cos 2 x = \ 4 \ cos 2 x ; sin 2 a? = \ — J cos 2 x. 

40. 1 + cos a: = 2 cos 2 - ; 1 — cos x = 2 sin 2 -- 

— A 

„.. . a; , ^ /l — cos a; a? , _ /l 4- cos a; 

41. sin-=±^ ;cos-=±V-^- ; 



2 *1 



cos a: 
tan 



cos a; 

42. sin a; 4- sin y = 2 sin 1 (a; + #) cos \(x — y}. 

43. sin a; — sin y == 2 cos J (a; + t y) sin J (a; — y). 

44. cos x 4- cos y = 2 cos ^ (a; 4- y) cos \(x — y^). 

45. cos a; — cos y = — 2 sin J (a? 4- ?/) sin J (a; — «/). 

46. -= - = -; Law of Sines. 

smi smB sin 6' 

47. a 2 — b 2 + c 2 — 2 be cos ^L ; Law of Cosines. 

FOEMULAS FROM ANALYTIC GEOMETRY 



48. d = V(a: 1 — x 2 ) 2 4- {jj\ — y 2 ) 2 » distance between points 
On #1) and (a- 2 , # 2 ). . 

49. d = — - — - ; distance from line 

± -J A 2 + B 2 Ax + By+ C= to (x v y{). 

50. x = -^ r-2, 2/ = ^ -p2 ; (^ ?/) is the point dividing the 

line -P 1 P 2 i n ^ ne rat i° ^- 



270 THEORETICAL MECHANICS 

51. x — x + x* , y — y + y f ; transforming to new origin (# , y ~). 

52. x = x' cos — y' sin 6, y = x' sin + y r cos 0; transforming 
to new axes making the angle with old. 

53. x = p cos 6, y = p sin ; transforming from rectangular to 
polar coordinates. 



54. /o = V# 2 + ?/ 2 , = arc tan — ; transforming from polar to 
rectangular coordinates. 

55. Different forms of equation of a straight line : 

, n y - V\ Vi - ¥i , ... 

(a) = , two-pomt form ; 

tAs tJU-t *Asn *Ay-t 

(b) - + \ == 1, intercept form ; 

(<?) y — y x = m(x — a^), slope-point form ; 

(d) y = m^ + 5, slope-intercept form ; 

(e) # cos cc -f y sin a = p, normal form ; 
(/) Ax + ify + 0= 0, general form. 

56. Distance from the line x cos ct -f- ?/ sin a— p = to the 
point (a^, y : ) = x 1 cos ot + y x sin a— p. 

57. tan = ^ — ; angle between two lines whose slopes are 

, 1 + m-, m 

m x and m 2 . x * 

m 1 = m 2 when lines are parallel, 

and m 1 = when lines are perpendicular. 

m 2 

58. (x — a) 2 + (?/ — /3) 2 = r 2 ; equation of circle with center 
(a, /3) and radius r. 



59. d = m y/(x 1 — x 2 y + {y± — y 2 } 2 + (^i — ^) 2 » distance between 
points (a^, y r s^) and (z 2 , y 2 , z 2 ). 

60. d = ^5 + ^ + &! + 2> . distance from plane 

±V^i 2 + J g 2 + (7 2 

-4a; + ify + Cz + 2> = to point (a^, y x , 2^) . 

61. cos 2 a + cos 2 /3 -f- cos 2 7 = 1; a, /3, 7 being the direction 
angles of a line in space. 



COLLECTION OF FORMULAS 



271 



62. x cos a -f- y cos -f- z cos 7 ; projection of the line joining 
(0, 0, 0) and (#, y, z) upon a line whose direction angles are a, /3, 7. 

63. (x — a) 2 + (y — /3) 2 -f (2 — 7) 2 = r 2 ; equation of sphere 
with center (a, /3, 7) and radius r. 

FORMULAS FROM CALCULUS 

64. Radius of curvature, 
(a) Rectangular coordinates. 

dy} 2 ^ 
dx, 



d 2 y 
dx 2 


MSI'J 


"-'S +2 SJ 


m+mi 



(6) Polar coordinates. 

B = 
(<?) Parametric form. 

dx d 2 y _ di£ d 2 x 
did^~ ~~dt~d& 

65. Plane area. 

(a) Rectangular coordinates. 

A = / ydx =11 dydx. 
(5) Polar coordinates. 

A=\Cp 2 de= C CpdpdO 

66. Length of arc. 

(a) Rectangular coordinates. 

(5) Polar coordinates. 



dx 



-JIGM*- 



']': 



= 7r I y 2 dx. 



272 THEORETICAL MECHANICS 

67. Volume of solid of revolution about the X-axis. 

V- 

68. Area of surface of revolution about the X-axis. 
A^jycU = 2,/„[l + (|) 2 ]^ = 2 *f*[($+lj«. 

69. Area of any surface, z =f(x, y). 



//[- 



if 



dydx. 



70. Volume of any solid. 

V= j I I dzdydx. 

DIFFERENTIAL EQUATIONS 

71. The differential equation of Harmonic Motion. 

air 



The general solution may be written in the following forms : 
(a) x = Cl e ktV ^ + c 2 e- k/z i, 

(6) x = Acoskt + BsmJct. 

We give to (5) another form, thus : 

Draw a right triangle with sides A and B. Since A and B are 

arbitrary constants, this right triangle is ar- 

. bitrary, and hence also the hypotenuse C and 

the angle ft. Now, 

A = C sin ft 5= (7 cos A 

and substitution in (b) gives 

x — (7(sin /3 cos Atf + cos /3 sin &£), or, 




CO 



x= O sin (H + /3). 



7T 



If in (<?) we write for /3, ft* + — , we obtain 

A 



IT 



0*) 



2:= <7sin(fa + /3' + £), or 
a; = (7 cos (kt + fi 1 ). 



COLLECTION OF FORMULAS 273 

In these formulas c v c v A, B, C, /3, /3' denote arbitrary 
constants. 



72. 






d 2 x 
dt 2 ' 


-k 2 x-. 


= 0. 


The 


general solution 


is 












x - 


= c x e kt 


+ c 2 e 


-» 



73. The differential equation of Damped Vibration. 

at 2, at 

The general solution is 



x = e'^XA cos V& 2 — fjfit+B sin V& 2 — /jl 2 £), or 



x = Ce'^ cos ( Vk 2 - fji 2 t + /3). 

74. The differential equation of harmonic motion with a con- 
stant disturbing force. 

ax. 7 o 

__ -|- Jc 2 x = c. 

at 2. 



The general solution is 
x = A cos 

x = C sin (Jet + /3) + 



# = J. cos kt + B sin &£ -+- — , or 

& 2 

c 



(a) — - 4- k 2 x = L cos w£ + ilfsin nt, where n^k. 

dt 2 



75. The differential equation of Forced Vibration. 

o) &+**=*< 

The general solution is 

x = A cos kt + B sin Atf -f- — cos nt + — sin nt, 

k l — n 2 k l — n 2 

where A and B are arbitrary constants. 

( &) C -^ + k 2 x = L cos Jct + M sin &£. 

eft 2 

The general solution is 

a; = ^L cos kt-\- B sin &£ H £ sin &£ — — — t cos kt. 

2k 'Ik 



274 THEORETICAL MECHANICS 



FORMULAS EOR DIFFERENTIATION 

In these formulas, u, v, and w denote variable quantities which 
are functions of x. 

i *£=o. 

dx 

II — = 1. * 

dx 

in A (u + v - w) = ta + ^-i»>. 

dx dx dx dx 

IV A (cv) = c ^. 

dx dx 

v A (uv) = u ^ + V <^. 

dx dx dx 

vi A ^ lV2 ... Vn ) = ( V2V3 ... Vn) ^i + ( VlVs ... Vn) ^ + ... 

dx dx dx 

. / \ dv n 

+ (Vi0 2 — V n -l) -*' 

dx 
VII 

Vila 



VIII 
Villa 
VIII & 

IX 
IX a 

X 

Xa 



i-o«)= 


= ™*-i^. 


dx v 


dx 


A r x n) = 
dx v J 


= ^zx^- 1 . 




dx dx 

-y 2 


d_/u\_ 
dx\v) 




dw 


A(u\- 
dx \G ) 


dx 


c 




du 

dx 


dx \vj 




du 


^Q0gaV)-- 


n dx 
= log a e 


dx 


v 




dv 


d 


dx 


¥^ ogv) - 


V 


Jw 


- a v \og a -^ • 

dx 


-(*) = 


„du 


dx 


dx 



XI ^.(w) = W 1 ^ + logtt-w B ' 

dx dx dx 



COLLECTION OF FORMULAS 



275 



XII 
XIII 
XIV 

xv 

XVI 

XVII 

XVIII 



-^-(sinv)=cos^. 
dx dx 

-f-(cosv) = -sinv^. 
dx dx 

!~(tanv) = sec2u^. 
dx dx 

dx dx 

— (sec v) = sec u tan v — • 
dx dx 

— (esc v) = — CSC v cot V — . 
dx dx 

— (vers v)= sin v^- 
dx dx 



XIX 



XX 



XXI 





dv 


V" (arc sin v) = 


dx 


dx J 


VI -v 2 


— (arc cos v) = 
dx 


dv 
' dx 


VI -v 2 




dv 


d , 

— (arc tan v) = 

dx y J 


dx 
1 + v* 



XXII 



XXIII 



(arc cot v) = — 



dv 
dx 



1 + v 2 



dv 
dx 



— (arc sec v) — 

dx vVv 2 -l 



dv 

XXIV A (arc esc ») = -^— 

dx uVv -2 - 1 

d_y 

XXV — (arc vers ») = — C • 

dx V2v- v 2 



XXVI 
XXVII 



dy_dydv 
dx dv dx 



y being a function of v. 



-^ = — , y being a function of a. 
dx dx 

dy 



276 THEORETICAL MECHANICS 

INTEGRALS FOR REFERENCE 

SOME ELEMENTARY FORMS 

1. | (du ±dv±dw ± •••)= \ du± \dv± \ dw ± 

2. \adv = a(dv. 4. (x n dx = ^—- + C, w=£l« 

3. jdf(x)=fjf(x)d£ = f(x)+C. 5. ^ = \ogx+a 

Forms containing Integral Powers of a + bx 

6. (^— = hog(a + bx)+C. 
J a + bx b 

7. f (a + bxydx = ( a + bx ^ n+1 + G, n =£ 1. 
J b(n + 1) . 

8. ( F(x, a + bx)dx. Try one of the substitutions, = a + foe, £2 = a + 6x. 

9. f-^L=l[a + foc-«log(a + fo*:)] + a 
J a + &x b 2 

10. f _^L = I[i(a + te) 2 -2a(a + &x) + a 2 log (a + 6x)] + (7. 
J a + foe b 8 

11. f__^ — = _liog«±^+a. 

-'x(a-ffoe) a x 

12 - f „*'„ =-^ + Alog^+^ + C. 
J x 2 (a + foc) «x a 2 x 

13. f_^ — = l[iog(a + fo*0 + — — l + C 
J (a + foe) 2 fo 2 L a + fod 

14. r ^ fa = l[ a + bx _ 2 a i g(q + foe) — 1 + C. 

J(a + fo*:) 2 ' fo*L ^ oV ' a + 6xj 

15. f ** = 1 Uog a + hx +C. 

J x(a + bx) 2 a(a + foe) a 2 x 

i6. r «** = ir — l_ + « i + o. 

J (a + 6x) 3 & 2 L a + bx 2(a + foe) 2 J 

Forms containing a 2 + x 2 , a 2 — x 2 , a + 6x", a + foe 2 

17. f dx = *tan-ig+ C; f -^- = tan-i x + C. 
J a 2 + x 2 a a J 1 + x 2 

is. r ^ = -iio g «±« + c; r dy = j-io g ^i« + -a 

J a 2 — x 2 2 a " a — x «/ x 2 — a' 2 2 a x + a 

19, f f '1 i = "7= tan_1 x \/- + C, when a > and b > 0. 
Ja + 6x 2 Va6 >a 



COLLECTION OF FORMULAS 277 

20. f dx = JL]ogg±Jg +a 
JcP-bW 2a6 °a-bx 

21. fx W! (a + 6x n )*>dx 

= xm-n +H g + to.)^i _ a(m-n + l) f m _„ (a ^ fc 
b(np + m + 1) b(np + m + 1) J 

22. f x™(a + 6x"> dx = xm+1 ( a + &x *) p + ®M (Wa + bx»)p-i dx. 

J np + m + 1 ?ip + m + 1 J 

23. f *= 

J x m (a + 6x n )i> 

1 (m — 7i + np — 1)6 f dx 

(m — l)ax m ~ 1 (« + bx n )p~ l (m - l)a J x m ~ n (a + 6x n )*> ' 

24. f *5 

J x m (a + bx n )p 



m — n -f n/j — 1 f dx_ 

5?i(p — 1) J X m (( 



an(i> — l)x m ~\a + 6x n ) / '~ 1 aw(.p — 1) J x m (a + 6x w ) p_1 

(a + bx n )p+ 1 _ 6(m - n — up — 1) C (a + 6a 
a(m — l)x'" _1 a(»i — 1) J x w ~ 

(a + 6x w )p anp f (a + bx 

(np — th + l^x™- 1 up — «i + 1 J r 



25 C (a 4- 6x n )? dx _ (a + 6x w )^+ 1 6(m - n — np - 1) f (a + bx n )p dx 
J x m a(m — l)x m - 1 a(m — l) J 

26 C (a + 6x n > c?x _ (a + 6x w )p anp f (a + 6x w )p- 1 dx 
J x" 1 (np — m + l)x m_1 np — m + 1 J 



J (a + 6x n )^ ~" 6(w — Tip + l)(a + 6x")^ _1 6(m - np + 1) J (a + bx n )r> 

2 q f x m dx x m+1 m + n — tijp + 1 f x^dx 

J (a + 6x n )*> ~~ anQi — l)(a + 6x n )p- 1 an(p - 1) J (a + 6x n ) p_1 

29. f d^_ = 1 J x + (2n _ 3) f dx 1 

J (a 2 + x 2 ) w 2(n - l)a 2 L(a 2 + x 2 )"- 1 y J (a 2 + a 2 )"- 1 J 

30. f dx = * T * 4-(2n-3)f - dx 1- 

J (a + bx*)" 2(n - l)a L(a + 6x 2 )»-i v ' J (a + 6X 2 )"- 1 .] 

31 r xdx = IC dz whereg==a . 2< 

J(a + 6x 2 )" 2J(a + 6^)« 

32 f _j^fo__ n^ , 1 _ C dx 

J (a + 6x 2 )"~2 6(n-l)(a + 6x 2 )' i - 1 2 6(n - 1) J (a + 6x 2 )"- 1 ' * 

33. C ^ = J-log %n + a 

J x(a + 6x n ) an a + 6x w 

34 f ^ —If dx - f c?x _ 

J x 2 (a + 6x 2 ) n ~ a J x\a + 6X 2 )"- 1 a J (a + 6x 2 )" ' 

35. r^^_ = J_logfx 2 + «Ua 37. f ^_ == J_log_^_+a 

Ja + 6x 2 26 V &/ Jx(a + 6x 2 ) 2 a a + 6x 2 

36 r x 2 dx _x a C dx 3g f dx _ 1 6 f dx 

J a + 6x 2 _ 6 6 J a + 6x 2 J x 2 (a + 6x 2 ) ax a J a + 6x 2 

39 C dx x J_ r dx 

J (a + 6x 2 ) 2 ~ 2 a(a + 6x 2 ) 2a J a + 6x 2 ' 



278 



THEOBETICAL MECHANICS 



Forms containing Va + bx 



40. 
41. 

42. 

43. 
44. 
45. 
46. 

47. 

48. 
49. 
50. 

51. 
52. 
53. 

54. 

55. 

56. 



(xV^TTxdx = - 2 (' 2a - s ^ v (^ b ^ 3 + c. 

J 15 6 2 

J 105 6 3 

I: 

f x 2 cfa _ 
*^ Va + 6x 



Va + bx 3 62 



2(8a 2 -4a6x + 3 5 2 x 2 ) v ^ T ^ 
15 63 



j <** = JLi og ^ + ^-v « + 



x Va + &x Va Va + &x + Va 
da; 2 



xVa + 6x V— a 

a"x — Va + frx b 



x 2 Va + 6x 



ax 



Va + &xax 



TVa_^ 
J x 



= 2 Va + bx + 



-i A /«±M + (7,fora<0. 

* — a 

_&_ r ax 

2 a./ jcVa + bx 

a ( dx . 

^ x Va + 6x 



Forms containing Vx 2 -f a' 2 
f (x 2 + arfdx = - Vx 2 + a 2 + — log (x + Vx 2 + a 2 ) + (7. 

«/ A A 

f (x 2 + a 2 )^ ax = - (2 x 2 + 5 a 2 ) Vx 2 + a 2 + — log (x + Vx 2 + a 2 ) + C. 
./ 8 8 

f (x 2 + a 2 ) 2 dx = x ( x * + fl ' 2) 2 + -ggL f(x 2 + a 2 ) 2_1 dx. 
J » + 1 n + 1 J 

w+2 

Tx(x 2 + a 2 ) 2 dx = ^ 2 + ^ 2 + C. 
J n + 2 

f x 2 (x 2 + a 2 )*dx = | (2 x 2 + a 2 ) Vx 2 + a 2 - -log (x + Vx 2 + a 2 ) + C 
•J 8 8 

h. 



dx 



(x 2 + a 2 ) 
dx 



= log (x + Vx 2 + a 2 ) 4 G. 



f 

(x 2 + a 2 )^ a 2 Vx 2 + a 2 

r xdx 



+ o. 



Vx 2 + a 2 + a 



(x 2 + a 2 ) 2 



J- 



x 2 ax 



(x 2 + a 2 ) 



T =| Vx 2 + a 2 - |- log (x + Vx 2 + a 2 ) + G. 

^ A A 



COLLECTION OF FORMULAS 279 

57. C — x2dx ^ = x + log ( X + Vx 2 + a 2 ) + C. 

J ( x -2 + a 2)l vV 2 + a' 2 

58. (* — — — = -\o g g + a 

x(x 2 + a 2 )* a a +Vx 2 + a 2 



eZx _ V x 2 + a 2 

a 2 x 



59. r — ^_ 



x 2 (x 2 4- a 2 )* 



60. I - = - V * 2 + <* 2 + J-l pg g+V^ +jgg +a 



). f S 

6i. ffi^±i^=v5rjFP-aiog«+^±? + a 

J X X 



62. 



f(x 2 +a 2 )^dx Vx 2 + a 2 , , , . /-o-: — 5n , ~ 

J X 2 X 



Forms containing vx 2 — a 2 
I 

63. f (x 2 - d 2 Ydx = x Vx 2 - a 2 _^log (x + Vx 2 - a 2 ) + C. 

64. f (x 2 - a 2 )^x = § (2 x 2 - 5 a 2 ) Vx 2 - a 2 + ^ log (x + Vx 2 - a 2 ) + C. 
Jo 8 

65. f (x 2 - a*)hx = x ( x *-f) 2 _ JiaL C (X 2 + tff^ax. 

J n+l n+ I J J 

■ n+2 

66. f x(x 2 - tffdx = ( x2 ~ a ^ 2 + c. 
J n+2 

67. f x 2 (x 2 - a 2 ) ^dx = - (2 x 2 - « 2 ) Vx 2 "^ 2 - ^ log (x + Vx 2 - a 2 ) + C. 
./ 8 8 

68. f ^ — r = log (x + Vx 2 - a 2 ) + 0. 

J (x 2 -a 2 )* 

69. f ^ = x + C. 

J ( x 2_ rt 2)f a?Vx*-a 2 

to. f *<** , = v^r^+a 

J ( X 2 _ a -2)t 

71 I s 2 <*E ^ = | Vx^^ 2 + |i og (x + Vx 2 ^^ 2 ) + a 
(x 2 — a 2 ) 7 * 

72. f — g2(to = x + log (x + Vx 2 - a 2 ) + C. 

J ( X 2 _ a 2)f Vx 2 - a 2 

73. f (fa Isec-ig+C; f — ^= = «ri« + <7. 

x(x 2 -a 2 )? a a J xVx 2 -l 



280 THEORETICAL MECHANICS 



74 C dx - ^ x2 — a ' 2 | c 

^ x 2 (x 2 - arf a ' 2x 



75. f & = V^l=_^ + J_ se c-^ + a 

J X X 



77< f (x 2 - a p*dx = _ Vx 2 - aa + ^-^ a 

J x' 2 x 

Forms containing Va 2 — x 2 

78. f (a 2 - x 2 )*ax = - Va 2 - x 2 + — sin-i - + (7. 

J A A CL 

79. (* (a 2 - x 2 ) ^x = - (5 a 2 - 2 x 2 ) Va 2 - x 2 + — sinr* 5 + #. 

80. (V - x 2 ) 2 c?x = x ^ 2 ~ **>' + -g*2L (V - x^'dx. 

J W + 1 71 + 1 J 

n+2 

81. ( x(a 2 - x 2 ) 2 dx = - (a 2 -'* 2 ) 2 + a 
J . n + 2 

82. fx 2 (a 2 - x 2 )^ c?x = - (2 x 2 - a 2 ) \/a 2 - x 2 + — sin" 1 - + 0. 
J 8 8 a 

83. f ^ — r = sin-i^; f dx =wa'^x. 

J ( a 2 _ aj2)i a J Vl - x 2 

84. f — — - = x + C. 85. f — xdx = - v/«2 _ X 2 + a 

J ( a 2 _ £2)1 a 2 Va 2 - x 2 J ( a 2 _ X 2)i 

g6 _ f — x^x__ = _ x v ^2^^2 + ^L 2 s in-i * + C. 

J (a 2 -x^ 2 2 a 

87. f — x 2 dx _ = _ _ x _ gin _ 1 x + a 

J ( a 2 _ ^2)1 Va 2 - x 2 « 

g8 C x«dx x^± vwzr¥2 + (m - l)a 2 C x"*- 2 ^ 

J ( a 2_ x 2)l ™ »» J ( a 2_ x 2) 1 2 

89. f *5 = ilcw. 



J 



x(a 2 -x 2 )^ a a+Va 2 -x 2 



90. f *? -=- ^ a2 ~ x2 +g 

x 2 (a 2 -x 2 )* a% 

9i. f *5 = _^ZEZ + J_i g » +q 

«/ ..« / -.o ..on * 2 a 2 x 2 2 a 3 „ j_ ,/„2 _ «z 



x 3 (a 2 - x 2 ) 1 



COLLECTION OF FORMULAS 281 



92. ( ( a2 -^ l dx=Vtf=&-a\o g a + V « i - x2 +a 

J X X 



^ ~ x ^ dx =- Va2 - x2 -sin-^+ a 



J x 2 



Forms containing V2 ax — x 2 , V2 ax -f x 2 

94. f V2ax-x 2 ax = ?—-^ V2 ax - x 2 + — vers"* - + C. 
J 2 2 a 

95. f dx = vers-i ? ; f (fa = vers"* x + C. 
•^ V2 ax — x 2 a J V2 x — x 2 

96. fx-V2ax-x 2 dx =- ^H 2 ^-* 2 ) 1 (2m + l)a f x «-iV2ax - x 2 dx. 
J »i + 2 ' ?« + 2 J 

gy C dx _ V2 ax — x 2 , m — 1 T dx 

J x m W9.nv — ^2 (2 m— l)ax m ' (2wi — l)a J vm -\~/: 



ax — x* 



I r x TO dx _ x w - 1 V2ax-x- (2m-l)a f 



V2 ax - x 2 »» w ^ V2 ax - x 2 



/' V2 ax -x 2 dx= (2 ax-x 2 ) s m - 3 f V2 ax - x 2 

J x m (2m — S)ax m (2m — 3)aJ x"'- 1 

00. fx V2 ax - x 2 dx = - 3 ff ' 2 + ax ~ 2 x2 V2 ax - x 2 + — vers-* g ■ 
J 6 2 a 

01. f dx = _ V2ax-x 2 | a 

J xV2 ax — x* ax 

Q2 C xdx = _ V2 ax - x 2 + a vers" 1 - + C. 
^ V2 ax — x 2 a 

03. f * 2 ^ = _^±3a V2qa ._ a , 2+ 3 a . 2vers . 1 g +a 

J V2 ax - x 2 2 2 a 



04. 



C V2 ax - x 2 dx = ^2 ax - x 2 + a vers-i- + C. 
./ x a 



rV2ax-x^ a , = _ 2V2«x-: f * _ vers . 1 . a 
J x 2 x a 



0& C^^^'dx^-^^-^+C. 
J x 3 3 ax 3 

07. f * -= - y " a +0. 

(2 ax - x 2 )^ a 2 V2ax-x 2 

08. f — = x + a 

(2 ax - x 2 ) * «^ 2 «* ~ ^ 2 



09. 



( F(x, V2 ax — x 2 )dx = ( F(z + a, Va 2 - z 2 )^, where = x - a. 



282 THEORETICAL MECHANICS 

HO. ( dx log (x + a + V2 ax + x 2 ) + (7. 

■^ V2 ax + x 2 

111. \ F(x, V2 ax + x 2 )dx = \ F(z — a, Vs' 2 — a' 2 )ds:, where 2 = as + a. 

Forms containing a + bx± ex 2 

112. f ^ = 2 tan-i 2cx + & + C, when 6 2 < 4 ac . 

J a + 5x + ex 2 V4 ac - 6 2 V4 ac - 6 2 



113. f *5 = 1 ipg^. + ^- ^-^ +a. when5 2 >4ac. 

J a + &X + CX 2 a/7)2'_ 4 r/y. 



+ bx + ex 2 V& 2 - 4 ac 2 ex + 6 + V& 2 - 4 ac 

r * = 1 io g 

J a + bx - ex 2 W h 2 , 4 „,. 



^ = 1 _ log V6 2 + 4ac + 2cx-& | a 

6x - ex 2 V6 2 + 4 ac V& 2 + 4 ac - 2 ex +b 

115. f ^ — = J-l a(;2cx + 5 + 2VcVa + 5x4-cx 2 )+a. 
* Va + &x + ex 2 Vc 

116. f Va + bx + cx 2 dx 



2 ex + 6 



Va + bx + ex 2 - - — — log (2cx + b + 2VcVa + &x + cx 2 ) + C. 



4c 8c2 



117. f— 



<fc _ J_ £in _ 1 2 ex -ft 

Va + 6x — ex 2 Ve V& 2 + 4 ac 



118. f Va + 6x - cx 2 dx = 2cx ~ h Va + fcx-cx 2 + &2+4ac sin-i 2cx ~ 5 + C. 
J 4c 8,1 Vfc 2 + 4ac 



119. f *<** = ^ + ^ + ^ 2 _ -^io gr 2ex + 5 + 2Vc-V a + &x+cx 2 ) + a 

•^ Va + 6x + ex 2 c oi 



120 f ax^c _ Vg + bx — ex 2 5 . , 2 ex — 5 ^ 

•^ Va + bx — ex 2 c 2 c^ v/ ^ >2 + 4 ac 



Other Algebraic Forms 

121. (J*L±*dx = V(a + x)(6 + x) + (a - 6) log ( Va + a + V&Tx) + <7. 
J \& + x 

122. f J^l^ ax = Vfa-x)(6 + x) + (a + 6) sin-i-v/^±^ + C. 
J^&+x ia-\-b 



123. f J f Aii^ dx = - V(a+x)(&-x) - (a + 6) sin-i-v/^-^ + C. 
J *& — x 'a + 6 

124 



. f ^ 1 -^- X dx = - Vl - x 2 + sin-ix + C. 



125. f ^ = 2 sin- W/^ 1 ^ + <?• 

J V(x - a) ()8 - x) */3-a 



COLLECTION OF FORMULAS 283 

Exponential and Trigonometric Forms 

126. f a x dx = -^— + C. 10Q r ■ * . n 
J log a 129 - \ Sln %dx = — cos x + C. 

127. (e*ax = &+ C. 

130. ( cos xdx = sin x + C. 

128. (<?«<7x=— + a 

131. ( tan xdx = log sec x = — log cos x + C. 

132. ( cot xdx = log sin x+ C. 

133. f sec xdx = C-^L = log(sec x + tan x) = log tan ( - + -"\ + C. 
J ^ cosx V"* 2/ 

134. (cosec acta = f -^- = log(cosec x — cot a:) = log tan f + C. 
J J sin x 2 

135. ( sec 2 xdx = tan x + (7. 138. ( cosec a: cot xdx = — cosec x+C. 

136. f cosec 2 xdx = - cot x + a 139 - Jsi^dx = * - - sin 2 x 4- C. 

137. f sec x tan xdx = sec x + C. 14 °- j*cos 2 xdx = | + | sin 2 x + G. 

141. f sin»xdx = - sinn ^cosx + n-1 f ^-2^. 
J n n J 

142. f cos»xdx = cosW " la; sin x + iLzi f cos*- 2 xdx. 

143 f jjg . 1 cosx , n — 2 f dx 

J sin w x w — 1 sin^-ix w — 1 J sin n ~ 2 x 

144 f (7a; __ 1 sin x . n — 2 C dx 

J COS«X 71 — 1 cos'*-^ w — 1 J COS"~ 2 X 

145. f cos"x sin»xdx = cosm ~ lx sin^x + to-J. f cosm _ 2a . s i n n^ # 
J m + n to + « J 

146. \ cos m x sin"xdx = - — — =-^-5 + \ cos TO x sin M - 2 xdx. 

J m + n m + nJ 

147 f <fa 1 1 , to + ii-2 r dx 

J sin m x cos n x n — 1 sin m-1 x cos n_1 x n — 1 J sin m x cos w -' 2 x 

148 C dx 1 , to 4- w — 2 f dx 



f dx 1 1 m + n-2 T 

J sin TO x cos n x to — 1 sin'"- 1 :*; cos' 1- ^ to — 1 J 



to — 1 sin m - a x cos' 1- ^ to — 1 J sin OT - 2 x cos n x 



un f cos m xdx _ __ cos OT + 1 x _ to — w + 2 Tcos^dx 
J sin n x (n — l)sin n ~ 1 x n — 1 J sin«- 2 x 

-, cq f cos m .rdx _ cpgw-ij. to — 1 rcos m -' 2 xdx 

J sin n x Cto — n^smP-kc to — w J 



sin"x (to — w)sin"- 1 x m — nJ sin n x 

os n+1 a 
w + 1 



151. Jsin x cos«xdx = - cosn+ ^ 4 



284 THEORETICAL MECHANICS 

152. fsin n a; cos xdx = smn+lx + <?. 
J n + 1 

153. (tan"xdx = tann ~ lx - f tan«- 2 adx + C. 
J n — 1 J 

154. f cot»xda; = - cotw ~ lx - (*cot"- 2 x(fa + G. 
J n—l J 

-ikk fc.:„ mn . c^ ™,7~ sin(m + n)x . sm(m — ri)x . ^ 

155. I sm rox sin nxdx — ^ ! — J - — | ^- — f- C7. 

J 2(m + ri) 2(w — «) 

1KC f„„„ „ *„„ ~j sin(m + n)x . sin(m — n)x , n 

156. \ cos mx cos nxdx = — * ! — J - — I ) l — f- G. 

J 2(m + n) 2(m — n) 

157. f sin mx cos nxdx = - cos ^ + n )* - cos(m = n > x + C. 
J 2(m + rc) 2(w-n) 

158. f -*5 = 2 tan-i ( J«E? tan ^ + O, when a > b. 

J a + b cosx Va 2 -& 2 ^a + 6 V 

y/b — a tan ? + V& + a 

159. ( — rr^ = , lo S — ^r h C, when a < 5. 

Ja + 6cosx V6 2 -« 2 V&^tan^-VM^ 

J a tan - + b 

^f. = — tan-i— =J=-+C, when a > 6. 
a + 6 sm x Va 2 - 6 2 vV - & 2 

a tan - + 6 - vV - a 2 

161. f ,f. = y 1 log + O, when a<6. 

Ja + 6sinx V & 2 - a 2 a tan* + 5+VP^ 

162. f ** = 1 tan-i ( *****) + G. 

J a 2 cos 2 x + 6 2 sin 2 x a b \ a ) 

163. f e» sin nxdx = e ^ a sin nx ~ n cos wa; ) + O ; 
jV sin xdx = eH^x^cosx) + ^ 

164. (V*cos wade = e °* (n shl ** + a cos nx KC; 
J a 2 + w 2 

jV cos xeZ* = ^(Binx + cosx) + c 

C owe 

165. \xe ax dx = — (ax-l)+ C. 
J a 2 

166. (x n e ax dx - x ^— - - [x n -^e a *dx. 
J a a J 

167. (d mx x n dx = -^—^ — (a mx x n ~ l dx. 

J m log a m log a J 



68. 



COLLECTION OF FORMULAS 

a x . log a C a x dx 

(m — l)x TO_1 mi — lJx™- 1 



285 



C a x di 
J x m 

69. (V COS «xdx = enoosn^xCacosx + nsmx) n(n-^ C ax co ^, xdx _ 
J a 2 •+- w' 2 a 2 + n? J 

r at" 1-1 m(vii I s ) C 

70. 1 x m cos axdx = — — (ax sm ax + m cos ax) ^ — - — J - \ x m ~ 2 cos axdx. 

J a 2 a 2 J 

Logarithmic Forms 

71. i log xdx = x log x — x + (7. 

72. rJ^=log(logx) + logx + llog 2 x+.... 
J log x 2 2 

73. r_^_=i g(log«)+a 

J x log X 

74. f x« log xdx = x«+i P^ - — ^1 + C. 
J Lw+1 (w + l) 2 J 

75. f e «log xdx = eaxl °Z x - 1 f^cfa. 
J a a./ a; 

76. f x™ log" xdx = %m+1 log* 1 x — f x™ log"" 1 xax. 

J m + 1 m + 1 J 

x m+1 . m + 1 C x m dx 

log"x (w ~ 1) log n_1 x n — 1 J log n - 1 x 



77. f " 



INDEX 



The numbers refer to the pages 



Acceleration, 45, 69 ; tangential and nor- 
mal, 80 
Angle of friction, 150 
Angular velocity, 84 ; momentum, 133 
Areal velocity, 134, 164 
Attraction of spheres, 214 
Atwood's machine, 139 
Axioms on force action, 100 

Bounding parabola, 144 

Catenary, 260 

Center of gravity, 1, 2, 5 ; of combined 

solids, 14 ; of wires, 17 
Center of moments, 131 
Central field of force, 163 
Centrifugal force, 118 
Centrodes, 229 
Circular motion, 87 
Coefficient of elasticity, 127 ; of friction, 

150, 255 
Composition of harmonic motions, 193, 

196 
Compound pendulum, 237 
Concurrent forces, 100 
Conservation of energy, 206 ; of moment 

of momentum, 221 
Conservative field of force, 205 
Constant field of force, 137 
Constrained motion, 116, 148 
Coplanar forces, 245 
Cord, flexible, 258 
Cycloid, motion of a particle on a, 158 

202 



Damped vibration, 51. 

Damping factor, 202 

Dimensions of units, 2, 8, 22, 29, 44, 

92, 111, 113, 127 
Distance-time diagram, 46 
Discussion of rectilinear motion , 46 

plane motion, 70 



46. 



of 



Dynamic pressure, 116 
Dynamics, 42 

Ellipse of inertia, 37 

Energy, kinetic, 113, 232 ; potential, 206 

Energy equation, 114, 135 

Epoch, 189 

Equation of motion, 42, 64 

Equilibrium of forces, 245 ; of flexible 

cords, 258 
Equipotential lines, 207 

Field of force, 137 ; central, 163 ; har- 
monic, 186, 198 ; conservative, 205 

Force, 90 ; units of, 92 ; resultant, 95 ; 
concurrent, 100 ; centrifugal, 118 ; mo- 
ments, 130 ; of restitution, 188 ; lines 
of, 207 

Force equations, 93, 101, 105, 135 

Forced vibrations, 196 

Formulas in dynamics of a particle, 135 

Frequency of vibration, 189 

Friction, 150, 213, 254 

Fundamental equations of motion, of 
particles, 134 ; of rigid body, 236 

Gravitation, universal, 183 
Gravity, motion under, 139, 142; work 
done by, 224 

Harmonic field of force, 186 
Harmonic motion, 49, 159, 188 
Height of projectile, 143 
Hooke's law, 112 

Impact, 127 

Impulse, 125 ; equation, 126, 136 

Inclined plane, 149 

Inertia ellipse, 37 

Instantaneous axis, 228 

Intrinsic force equations, 105, 135 



28^ 



288 



INDEX 



Kepler, 182 
Kinematics, 42, 55 
Kinetic energy, 113, 232 
Kinetics, 42, 90 

Lamina, 25 

Law of areas, 165 ; of universal gravita- 
tion, 183 

Laws of friction, 150, 255 ; of motion, 
90, 117 

Lever arm, 131 

Line of quickest descent, 152 

Lines of force, 207 

Lissajous 1 curves, 199 

Mass-center, motion of, 219, 232 
Moment equation, 132, 136, 220, 234 
Moment of area, 1 ; of mass, 7, 9 ; of in- 
ertia, 21, 24 ; of momentum, 132 
Momentum, 90 

Motion in a prescribed path, 76 
Motion, types of, 226 

Newton, 86, 90, 117, 127, 182, 213 
Newtonian potential, 213 
Non-conservative forces, 213 

Orbit, 163 ; differential equation of, 170 

Pappus, theorems of, 18 

Pendulum, simple, 154 ; cycloidal, 158 ; 

seconds, 160; compound, 237 
Period of vibration, 50 
Phase, 189 

Planetary motion, 182 
Polar coordinates, 83, 105, 135 



Potential function, 167, 205; energy, 

206 ; newtonian, 213 
Power, 122 
Product of inertia, 38 
Projectile, 142 

Eadial velocity, 84 
Kadius of gyration, 23 
Range of projectile, 143 
Resisting medium, 95, 200 
Restitution, impulse of, 127 
Rigid body, 226 
Rotation, 87, 227 
Kouth's rules, 34 

Screw motion, 229 
Seconds pendulum, 160 
Speed, 44, 66 
Stability, 262 
Static pressure, 118 
Symmetry, 3 

System of bodies, 238 ; of particles, 36, 
217 

Tautochronous curve, 160 
Translation, 226 

Uniplanar motion, 227 
Units, fundamental and derived, 2 ; of 
force, 92 ; of work, 123 ; of power, 123 

Vectors, 55 ; addition of, 56 ; subtraction 

of, 57 ; resolution of, 58 
Velocity, 43, 65, 84 

Weight, 91 
Work, 109, 231 



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